Prime ideals and examples of them

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So the question states that the intersection of two prime ideals is always a prime ideal. Well this is false but I need an example to counter it. I looked online and found one "For example, inside Z, 2Z and 3Z are prime, but there intersection, 6Z is not prime"

so I just need some explanation to what a prime ideal is and how you can determine that an ideal is prime. The definition I know of is

Let $R$ be a comm. ring with identity. An ideal P is prime iff $R neq P$ and whenever $bc in P$ then $b in P$ or $c in P$

I dont know how to apply this definition to the example above.

abstract-algebra

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edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

• 
  
  
  

  
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  Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
  – Qiaochu Yuan
  Feb 26 '11 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
  – beroal
  Feb 26 '11 at 20:41
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

6

So the question states that the intersection of two prime ideals is always a prime ideal. Well this is false but I need an example to counter it. I looked online and found one "For example, inside Z, 2Z and 3Z are prime, but there intersection, 6Z is not prime"

so I just need some explanation to what a prime ideal is and how you can determine that an ideal is prime. The definition I know of is

Let $R$ be a comm. ring with identity. An ideal P is prime iff $R neq P$ and whenever $bc in P$ then $b in P$ or $c in P$

I dont know how to apply this definition to the example above.

abstract-algebra

share|cite|improve this question

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
  – Qiaochu Yuan
  Feb 26 '11 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
  – beroal
  Feb 26 '11 at 20:41
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

6

up vote
6
down vote

favorite

6

6

So the question states that the intersection of two prime ideals is always a prime ideal. Well this is false but I need an example to counter it. I looked online and found one "For example, inside Z, 2Z and 3Z are prime, but there intersection, 6Z is not prime"

so I just need some explanation to what a prime ideal is and how you can determine that an ideal is prime. The definition I know of is

Let $R$ be a comm. ring with identity. An ideal P is prime iff $R neq P$ and whenever $bc in P$ then $b in P$ or $c in P$

I dont know how to apply this definition to the example above.

abstract-algebra

share|cite|improve this question

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

So the question states that the intersection of two prime ideals is always a prime ideal. Well this is false but I need an example to counter it. I looked online and found one "For example, inside Z, 2Z and 3Z are prime, but there intersection, 6Z is not prime"

so I just need some explanation to what a prime ideal is and how you can determine that an ideal is prime. The definition I know of is

Let $R$ be a comm. ring with identity. An ideal P is prime iff $R neq P$ and whenever $bc in P$ then $b in P$ or $c in P$

I dont know how to apply this definition to the example above.

abstract-algebra

share|cite|improve this question

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

share|cite|improve this question

share|cite|improve this question

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

edited Mar 8 '17 at 12:10

gymbvghjkgkjkhgfkl

1

1

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

asked Feb 26 '11 at 1:53

Tyler Hilton

1,37611848

1,37611848

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
  – Qiaochu Yuan
  Feb 26 '11 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
  – beroal
  Feb 26 '11 at 20:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
  – Qiaochu Yuan
  Feb 26 '11 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
  – beroal
  Feb 26 '11 at 20:41
  

  
  
  
  

4

4

Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
– Qiaochu Yuan
Feb 26 '11 at 2:06

Can you find b, c such that bc is divisible by 6 but such that neither b nor c is divisible by 6?
– Qiaochu Yuan
Feb 26 '11 at 2:06

Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
– beroal
Feb 26 '11 at 20:41

Counterexample. b=2, c=3. bc in 6Z and b not in 6Z and c not in 6Z. You do not really need a long answer.
– beroal
Feb 26 '11 at 20:41

add a comment | 

3 Answers
3

active

oldest

votes

up vote
21
down vote



accepted

I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.

Ideals, at least at first, are meant to generalize the notion of "is a multiple of" (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at this answer about normal subgroups later).

If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples,
$$nmathbbZ = (n) = ainmathbbZmid atext is a multiples of n.$$
What properties do the collections of "all multiples of a given number" have? Well:

• The collection always contains $0$.


• If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.


• If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.

In the integers, and also in many other rings (for example, $mathbbR[x]$, the polynomials over $mathbbR$), every collection that satisfies these three properties is in fact the collection of all multiples of some $ain R$. But there are other rings where this does not happen. For example, if you consider the ring $mathbbZ[x]$ of all polynomials with integer coefficients, you can take
$$I = p(x)inmathbbZ[x]mid p(0)text is even.$$
This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal.

Is this $I$ the collection of "all multiples of $a$" for some $ainmathbbZ[x]$? No. If there were such an $a$, then since $2in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $pm 1$ or $pm 2$ (the only elements of $mathbbZ[x]$ that divide $2$). It can't be either $1$ or $-1$, because "multiples of $pm 1$" is everything, and not everything is in our $I$. But $xin I$ as well, since evaluating $x$ at $0$ is even; and neither $2$ nor $-2$ divide $x$ in $mathbbZ[x]$. So even though $I$ is an ideal, it is not "all multiples of" someone. So the notion of "ideal", even though it starts up as "all multiples of" someone, is actually more general. This is the distinction between principal ideals (ideals which are "all multiples of $a$" for some $a$), and more general ideals (which need not be made up of "all multiples of $a$" for some $a$).

Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do "modulo arithmetic" as in the integers (working modulo $n$ is "really" working in $mathbbZ/(n)$) then the conditions you need on a collection are precisely the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do "modulo arithmetic". And modulo arithmetic is all about divisibility (after all, $aequiv bpmodn$ means that $n$ divides $a-b$).

So we want to also keep track of a few of the other special properties that some numbers have, and "translate" them into what they mean for ideals.

Prime numbers play a major role in divisibility issues in the integers. How does the "prime" property translate into the setting of ideals? A prime is a number $p$ such that:

• $pneq pm 1$; and


• If $p$ divides a product, then it divides at least one of the factors.

Okay, how does that translate into ideals? If you think of an ideal $I$ as "the collection of all multiples of some number" (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide everything; so we require the ideal to not be the entire ring, $Ineq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $abin I$, then either $ain I$ or $bin I$. So we define:

An ideal $I$ is a prime ideal if and only if $Ineq R$, and whenever $abin I$, either $ain I$ or $bin I$.

Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like "all multiples of $a$", and $J$ is sort of like "all multiples of $b$", then what is $Icap J$? All things that are multiples of both $a$ and $b$!

So, if $P$ and $Q$ are both prime ideals, would $Pcap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of two different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $mathbbZ$, precisely because $2$ is a prime number: if $abin(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime number), so either $ain(2)$ or $bin(2)$. Similarly with $(3)$. But $(2)cap(3)$ will be all numbers that are multiples of both $2$ and $3$; this corresponds to "all multiples of $6$", as we know from elementary number theory: $(2)cap(3)=(6)$. But $6$ is not a prime number, so there is no reason to expect $(6)$ to be a prime ideal. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.

Caveat. The analogy of ideals as "set of all multiples of something" works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $pneq 0$, $qneq 0$, $pneq pm q$, and $(p)subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=mathbbR[x,y]$, the ring of polynomials in two variables, both
beginalign*
(x) &= p(x,y)in Rmid p(0,y) = 0text for all y;\
(x,y) &= p(x,y)in Rmid p(0,0) = 0
endalign*
are ideals; clearly $(x)subseteq (x,y)$, $(x)neq (0)$, $(x,y)neq (0)$, and $(x)neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals.

So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples.

share|cite|improve this answer

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
  – Tyler Hilton
  Feb 27 '11 at 2:22
  

  
  
  
  

add a comment | 

up vote
2
down vote

An equivalent definition of prime ideals is the following:

An ideal I in a ring R is a prime ideal if the quotient ring R/I is an integral domain.

Proof of equivalence: Assume $I$ is a prime ideal in "your" definition. And assume $xy=0$ in $R/I$. This says precisely that $xy in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.

For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.

In $mathbbZ$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $mathbbZ$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)

share|cite|improve this answer

edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

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  In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
  – awllower
  Feb 26 '11 at 3:21
  
  

  
  
  
  

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By definition, $rm 6 mathbb Z $ isn't prime $rmiff$ there are $rm:b,c $ such that $rm bc in 6 mathbb Z, b,c notin 6 mathbb Z:.:$ Equivalently, $rm 6 | bc, 6 nmid b,c:. $ So it suffices to exhibit a proper factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an irreducible polynomial. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).

share|cite|improve this answer

edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

Number

1

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  What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
  – hhh
  Feb 13 '16 at 14:08
  
  

  
  
  
  


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  @hhh Yes, that's what UFD denotes in ring theory.
  – Number
  Jun 4 '17 at 16:18
  

  
  
  
  

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
21
down vote



accepted

I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.

Ideals, at least at first, are meant to generalize the notion of "is a multiple of" (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at this answer about normal subgroups later).

If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples,
$$nmathbbZ = (n) = ainmathbbZmid atext is a multiples of n.$$
What properties do the collections of "all multiples of a given number" have? Well:

• The collection always contains $0$.


• If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.


• If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.

In the integers, and also in many other rings (for example, $mathbbR[x]$, the polynomials over $mathbbR$), every collection that satisfies these three properties is in fact the collection of all multiples of some $ain R$. But there are other rings where this does not happen. For example, if you consider the ring $mathbbZ[x]$ of all polynomials with integer coefficients, you can take
$$I = p(x)inmathbbZ[x]mid p(0)text is even.$$
This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal.

Is this $I$ the collection of "all multiples of $a$" for some $ainmathbbZ[x]$? No. If there were such an $a$, then since $2in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $pm 1$ or $pm 2$ (the only elements of $mathbbZ[x]$ that divide $2$). It can't be either $1$ or $-1$, because "multiples of $pm 1$" is everything, and not everything is in our $I$. But $xin I$ as well, since evaluating $x$ at $0$ is even; and neither $2$ nor $-2$ divide $x$ in $mathbbZ[x]$. So even though $I$ is an ideal, it is not "all multiples of" someone. So the notion of "ideal", even though it starts up as "all multiples of" someone, is actually more general. This is the distinction between principal ideals (ideals which are "all multiples of $a$" for some $a$), and more general ideals (which need not be made up of "all multiples of $a$" for some $a$).

Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do "modulo arithmetic" as in the integers (working modulo $n$ is "really" working in $mathbbZ/(n)$) then the conditions you need on a collection are precisely the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do "modulo arithmetic". And modulo arithmetic is all about divisibility (after all, $aequiv bpmodn$ means that $n$ divides $a-b$).

So we want to also keep track of a few of the other special properties that some numbers have, and "translate" them into what they mean for ideals.

Prime numbers play a major role in divisibility issues in the integers. How does the "prime" property translate into the setting of ideals? A prime is a number $p$ such that:

• $pneq pm 1$; and


• If $p$ divides a product, then it divides at least one of the factors.

Okay, how does that translate into ideals? If you think of an ideal $I$ as "the collection of all multiples of some number" (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide everything; so we require the ideal to not be the entire ring, $Ineq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $abin I$, then either $ain I$ or $bin I$. So we define:

An ideal $I$ is a prime ideal if and only if $Ineq R$, and whenever $abin I$, either $ain I$ or $bin I$.

Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like "all multiples of $a$", and $J$ is sort of like "all multiples of $b$", then what is $Icap J$? All things that are multiples of both $a$ and $b$!

So, if $P$ and $Q$ are both prime ideals, would $Pcap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of two different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $mathbbZ$, precisely because $2$ is a prime number: if $abin(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime number), so either $ain(2)$ or $bin(2)$. Similarly with $(3)$. But $(2)cap(3)$ will be all numbers that are multiples of both $2$ and $3$; this corresponds to "all multiples of $6$", as we know from elementary number theory: $(2)cap(3)=(6)$. But $6$ is not a prime number, so there is no reason to expect $(6)$ to be a prime ideal. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.

Caveat. The analogy of ideals as "set of all multiples of something" works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $pneq 0$, $qneq 0$, $pneq pm q$, and $(p)subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=mathbbR[x,y]$, the ring of polynomials in two variables, both
beginalign*
(x) &= p(x,y)in Rmid p(0,y) = 0text for all y;\
(x,y) &= p(x,y)in Rmid p(0,0) = 0
endalign*
are ideals; clearly $(x)subseteq (x,y)$, $(x)neq (0)$, $(x,y)neq (0)$, and $(x)neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals.

So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples.

share|cite|improve this answer

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
  – Tyler Hilton
  Feb 27 '11 at 2:22
  

  
  
  
  

add a comment | 

up vote
21
down vote



accepted

I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.

Ideals, at least at first, are meant to generalize the notion of "is a multiple of" (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at this answer about normal subgroups later).

If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples,
$$nmathbbZ = (n) = ainmathbbZmid atext is a multiples of n.$$
What properties do the collections of "all multiples of a given number" have? Well:

• The collection always contains $0$.


• If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.


• If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.

In the integers, and also in many other rings (for example, $mathbbR[x]$, the polynomials over $mathbbR$), every collection that satisfies these three properties is in fact the collection of all multiples of some $ain R$. But there are other rings where this does not happen. For example, if you consider the ring $mathbbZ[x]$ of all polynomials with integer coefficients, you can take
$$I = p(x)inmathbbZ[x]mid p(0)text is even.$$
This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal.

Is this $I$ the collection of "all multiples of $a$" for some $ainmathbbZ[x]$? No. If there were such an $a$, then since $2in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $pm 1$ or $pm 2$ (the only elements of $mathbbZ[x]$ that divide $2$). It can't be either $1$ or $-1$, because "multiples of $pm 1$" is everything, and not everything is in our $I$. But $xin I$ as well, since evaluating $x$ at $0$ is even; and neither $2$ nor $-2$ divide $x$ in $mathbbZ[x]$. So even though $I$ is an ideal, it is not "all multiples of" someone. So the notion of "ideal", even though it starts up as "all multiples of" someone, is actually more general. This is the distinction between principal ideals (ideals which are "all multiples of $a$" for some $a$), and more general ideals (which need not be made up of "all multiples of $a$" for some $a$).

Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do "modulo arithmetic" as in the integers (working modulo $n$ is "really" working in $mathbbZ/(n)$) then the conditions you need on a collection are precisely the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do "modulo arithmetic". And modulo arithmetic is all about divisibility (after all, $aequiv bpmodn$ means that $n$ divides $a-b$).

So we want to also keep track of a few of the other special properties that some numbers have, and "translate" them into what they mean for ideals.

Prime numbers play a major role in divisibility issues in the integers. How does the "prime" property translate into the setting of ideals? A prime is a number $p$ such that:

• $pneq pm 1$; and


• If $p$ divides a product, then it divides at least one of the factors.

Okay, how does that translate into ideals? If you think of an ideal $I$ as "the collection of all multiples of some number" (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide everything; so we require the ideal to not be the entire ring, $Ineq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $abin I$, then either $ain I$ or $bin I$. So we define:

An ideal $I$ is a prime ideal if and only if $Ineq R$, and whenever $abin I$, either $ain I$ or $bin I$.

Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like "all multiples of $a$", and $J$ is sort of like "all multiples of $b$", then what is $Icap J$? All things that are multiples of both $a$ and $b$!

So, if $P$ and $Q$ are both prime ideals, would $Pcap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of two different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $mathbbZ$, precisely because $2$ is a prime number: if $abin(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime number), so either $ain(2)$ or $bin(2)$. Similarly with $(3)$. But $(2)cap(3)$ will be all numbers that are multiples of both $2$ and $3$; this corresponds to "all multiples of $6$", as we know from elementary number theory: $(2)cap(3)=(6)$. But $6$ is not a prime number, so there is no reason to expect $(6)$ to be a prime ideal. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.

Caveat. The analogy of ideals as "set of all multiples of something" works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $pneq 0$, $qneq 0$, $pneq pm q$, and $(p)subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=mathbbR[x,y]$, the ring of polynomials in two variables, both
beginalign*
(x) &= p(x,y)in Rmid p(0,y) = 0text for all y;\
(x,y) &= p(x,y)in Rmid p(0,0) = 0
endalign*
are ideals; clearly $(x)subseteq (x,y)$, $(x)neq (0)$, $(x,y)neq (0)$, and $(x)neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals.

So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples.

share|cite|improve this answer

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
  – Tyler Hilton
  Feb 27 '11 at 2:22
  

  
  
  
  

add a comment | 

up vote
21
down vote



accepted

up vote
21
down vote



accepted

I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.

Ideals, at least at first, are meant to generalize the notion of "is a multiple of" (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at this answer about normal subgroups later).

If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples,
$$nmathbbZ = (n) = ainmathbbZmid atext is a multiples of n.$$
What properties do the collections of "all multiples of a given number" have? Well:

• The collection always contains $0$.


• If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.


• If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.

In the integers, and also in many other rings (for example, $mathbbR[x]$, the polynomials over $mathbbR$), every collection that satisfies these three properties is in fact the collection of all multiples of some $ain R$. But there are other rings where this does not happen. For example, if you consider the ring $mathbbZ[x]$ of all polynomials with integer coefficients, you can take
$$I = p(x)inmathbbZ[x]mid p(0)text is even.$$
This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal.

Is this $I$ the collection of "all multiples of $a$" for some $ainmathbbZ[x]$? No. If there were such an $a$, then since $2in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $pm 1$ or $pm 2$ (the only elements of $mathbbZ[x]$ that divide $2$). It can't be either $1$ or $-1$, because "multiples of $pm 1$" is everything, and not everything is in our $I$. But $xin I$ as well, since evaluating $x$ at $0$ is even; and neither $2$ nor $-2$ divide $x$ in $mathbbZ[x]$. So even though $I$ is an ideal, it is not "all multiples of" someone. So the notion of "ideal", even though it starts up as "all multiples of" someone, is actually more general. This is the distinction between principal ideals (ideals which are "all multiples of $a$" for some $a$), and more general ideals (which need not be made up of "all multiples of $a$" for some $a$).

Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do "modulo arithmetic" as in the integers (working modulo $n$ is "really" working in $mathbbZ/(n)$) then the conditions you need on a collection are precisely the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do "modulo arithmetic". And modulo arithmetic is all about divisibility (after all, $aequiv bpmodn$ means that $n$ divides $a-b$).

So we want to also keep track of a few of the other special properties that some numbers have, and "translate" them into what they mean for ideals.

Prime numbers play a major role in divisibility issues in the integers. How does the "prime" property translate into the setting of ideals? A prime is a number $p$ such that:

• $pneq pm 1$; and


• If $p$ divides a product, then it divides at least one of the factors.

Okay, how does that translate into ideals? If you think of an ideal $I$ as "the collection of all multiples of some number" (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide everything; so we require the ideal to not be the entire ring, $Ineq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $abin I$, then either $ain I$ or $bin I$. So we define:

An ideal $I$ is a prime ideal if and only if $Ineq R$, and whenever $abin I$, either $ain I$ or $bin I$.

Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like "all multiples of $a$", and $J$ is sort of like "all multiples of $b$", then what is $Icap J$? All things that are multiples of both $a$ and $b$!

So, if $P$ and $Q$ are both prime ideals, would $Pcap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of two different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $mathbbZ$, precisely because $2$ is a prime number: if $abin(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime number), so either $ain(2)$ or $bin(2)$. Similarly with $(3)$. But $(2)cap(3)$ will be all numbers that are multiples of both $2$ and $3$; this corresponds to "all multiples of $6$", as we know from elementary number theory: $(2)cap(3)=(6)$. But $6$ is not a prime number, so there is no reason to expect $(6)$ to be a prime ideal. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.

Caveat. The analogy of ideals as "set of all multiples of something" works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $pneq 0$, $qneq 0$, $pneq pm q$, and $(p)subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=mathbbR[x,y]$, the ring of polynomials in two variables, both
beginalign*
(x) &= p(x,y)in Rmid p(0,y) = 0text for all y;\
(x,y) &= p(x,y)in Rmid p(0,0) = 0
endalign*
are ideals; clearly $(x)subseteq (x,y)$, $(x)neq (0)$, $(x,y)neq (0)$, and $(x)neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals.

So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples.

share|cite|improve this answer

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.

Ideals, at least at first, are meant to generalize the notion of "is a multiple of" (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at this answer about normal subgroups later).

If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples,
$$nmathbbZ = (n) = ainmathbbZmid atext is a multiples of n.$$
What properties do the collections of "all multiples of a given number" have? Well:

• The collection always contains $0$.


• If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.


• If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.

In the integers, and also in many other rings (for example, $mathbbR[x]$, the polynomials over $mathbbR$), every collection that satisfies these three properties is in fact the collection of all multiples of some $ain R$. But there are other rings where this does not happen. For example, if you consider the ring $mathbbZ[x]$ of all polynomials with integer coefficients, you can take
$$I = p(x)inmathbbZ[x]mid p(0)text is even.$$
This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal.

Is this $I$ the collection of "all multiples of $a$" for some $ainmathbbZ[x]$? No. If there were such an $a$, then since $2in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $pm 1$ or $pm 2$ (the only elements of $mathbbZ[x]$ that divide $2$). It can't be either $1$ or $-1$, because "multiples of $pm 1$" is everything, and not everything is in our $I$. But $xin I$ as well, since evaluating $x$ at $0$ is even; and neither $2$ nor $-2$ divide $x$ in $mathbbZ[x]$. So even though $I$ is an ideal, it is not "all multiples of" someone. So the notion of "ideal", even though it starts up as "all multiples of" someone, is actually more general. This is the distinction between principal ideals (ideals which are "all multiples of $a$" for some $a$), and more general ideals (which need not be made up of "all multiples of $a$" for some $a$).

Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do "modulo arithmetic" as in the integers (working modulo $n$ is "really" working in $mathbbZ/(n)$) then the conditions you need on a collection are precisely the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do "modulo arithmetic". And modulo arithmetic is all about divisibility (after all, $aequiv bpmodn$ means that $n$ divides $a-b$).

So we want to also keep track of a few of the other special properties that some numbers have, and "translate" them into what they mean for ideals.

Prime numbers play a major role in divisibility issues in the integers. How does the "prime" property translate into the setting of ideals? A prime is a number $p$ such that:

• $pneq pm 1$; and


• If $p$ divides a product, then it divides at least one of the factors.

Okay, how does that translate into ideals? If you think of an ideal $I$ as "the collection of all multiples of some number" (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide everything; so we require the ideal to not be the entire ring, $Ineq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $abin I$, then either $ain I$ or $bin I$. So we define:

An ideal $I$ is a prime ideal if and only if $Ineq R$, and whenever $abin I$, either $ain I$ or $bin I$.

Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like "all multiples of $a$", and $J$ is sort of like "all multiples of $b$", then what is $Icap J$? All things that are multiples of both $a$ and $b$!

So, if $P$ and $Q$ are both prime ideals, would $Pcap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of two different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $mathbbZ$, precisely because $2$ is a prime number: if $abin(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime number), so either $ain(2)$ or $bin(2)$. Similarly with $(3)$. But $(2)cap(3)$ will be all numbers that are multiples of both $2$ and $3$; this corresponds to "all multiples of $6$", as we know from elementary number theory: $(2)cap(3)=(6)$. But $6$ is not a prime number, so there is no reason to expect $(6)$ to be a prime ideal. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.

Caveat. The analogy of ideals as "set of all multiples of something" works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $pneq 0$, $qneq 0$, $pneq pm q$, and $(p)subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=mathbbR[x,y]$, the ring of polynomials in two variables, both
beginalign*
(x) &= p(x,y)in Rmid p(0,y) = 0text for all y;\
(x,y) &= p(x,y)in Rmid p(0,0) = 0
endalign*
are ideals; clearly $(x)subseteq (x,y)$, $(x)neq (0)$, $(x,y)neq (0)$, and $(x)neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals.

So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples.

share|cite|improve this answer

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 16:33

edited Aug 6 at 16:33

edited Aug 6 at 16:33

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

answered Feb 26 '11 at 20:06

Arturo Magidin

254k31568886

254k31568886

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
  – Tyler Hilton
  Feb 27 '11 at 2:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
  – Tyler Hilton
  Feb 27 '11 at 2:22
  

  
  
  
  

2

2

Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
– Tyler Hilton
Feb 27 '11 at 2:22

Thank you so much. You have no idea how amazing it is to read something that makes so much sense rather then read textbooks which just spit out definitions.
– Tyler Hilton
Feb 27 '11 at 2:22

add a comment | 

up vote
2
down vote

An equivalent definition of prime ideals is the following:

An ideal I in a ring R is a prime ideal if the quotient ring R/I is an integral domain.

Proof of equivalence: Assume $I$ is a prime ideal in "your" definition. And assume $xy=0$ in $R/I$. This says precisely that $xy in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.

For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.

In $mathbbZ$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $mathbbZ$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)

share|cite|improve this answer

edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
  – awllower
  Feb 26 '11 at 3:21
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

An equivalent definition of prime ideals is the following:

An ideal I in a ring R is a prime ideal if the quotient ring R/I is an integral domain.

Proof of equivalence: Assume $I$ is a prime ideal in "your" definition. And assume $xy=0$ in $R/I$. This says precisely that $xy in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.

For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.

In $mathbbZ$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $mathbbZ$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)

share|cite|improve this answer

edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
  – awllower
  Feb 26 '11 at 3:21
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

An equivalent definition of prime ideals is the following:

An ideal I in a ring R is a prime ideal if the quotient ring R/I is an integral domain.

Proof of equivalence: Assume $I$ is a prime ideal in "your" definition. And assume $xy=0$ in $R/I$. This says precisely that $xy in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.

For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.

In $mathbbZ$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $mathbbZ$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)

share|cite|improve this answer

edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

An equivalent definition of prime ideals is the following:

An ideal I in a ring R is a prime ideal if the quotient ring R/I is an integral domain.

Proof of equivalence: Assume $I$ is a prime ideal in "your" definition. And assume $xy=0$ in $R/I$. This says precisely that $xy in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.

For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.

In $mathbbZ$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $mathbbZ$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)

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edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

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share|cite|improve this answer

edited Feb 26 '11 at 3:01

edited Feb 26 '11 at 3:01

edited Feb 26 '11 at 3:01

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

answered Feb 26 '11 at 2:51

Fredrik Meyer

14.9k23563

14.9k23563

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  In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
  – awllower
  Feb 26 '11 at 3:21
  
  

  
  
  
  

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  In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
  – awllower
  Feb 26 '11 at 3:21
  
  

  
  
  
  

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1

In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
– awllower
Feb 26 '11 at 3:21

In the case of $mathbbZ$, it is a unique factorization domain, so the intersection of two mutually prime ideals, viewed as the lowest common multiple of them, is not prime at all. Since the notion of unique prime ideal factorization is the same as the Dedekind domains, it fits a wide range of fields, or rings.
– awllower
Feb 26 '11 at 3:21

add a comment | 

up vote
2
down vote

By definition, $rm 6 mathbb Z $ isn't prime $rmiff$ there are $rm:b,c $ such that $rm bc in 6 mathbb Z, b,c notin 6 mathbb Z:.:$ Equivalently, $rm 6 | bc, 6 nmid b,c:. $ So it suffices to exhibit a proper factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an irreducible polynomial. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).

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edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

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  What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
  – hhh
  Feb 13 '16 at 14:08
  
  

  
  
  
  


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  @hhh Yes, that's what UFD denotes in ring theory.
  – Number
  Jun 4 '17 at 16:18
  

  
  
  
  

add a comment | 

up vote
2
down vote

By definition, $rm 6 mathbb Z $ isn't prime $rmiff$ there are $rm:b,c $ such that $rm bc in 6 mathbb Z, b,c notin 6 mathbb Z:.:$ Equivalently, $rm 6 | bc, 6 nmid b,c:. $ So it suffices to exhibit a proper factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an irreducible polynomial. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).

share|cite|improve this answer

edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

Number

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
  – hhh
  Feb 13 '16 at 14:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @hhh Yes, that's what UFD denotes in ring theory.
  – Number
  Jun 4 '17 at 16:18
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

By definition, $rm 6 mathbb Z $ isn't prime $rmiff$ there are $rm:b,c $ such that $rm bc in 6 mathbb Z, b,c notin 6 mathbb Z:.:$ Equivalently, $rm 6 | bc, 6 nmid b,c:. $ So it suffices to exhibit a proper factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an irreducible polynomial. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).

share|cite|improve this answer

edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

Number

1

By definition, $rm 6 mathbb Z $ isn't prime $rmiff$ there are $rm:b,c $ such that $rm bc in 6 mathbb Z, b,c notin 6 mathbb Z:.:$ Equivalently, $rm 6 | bc, 6 nmid b,c:. $ So it suffices to exhibit a proper factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an irreducible polynomial. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).

share|cite|improve this answer

edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

Number

1

share|cite|improve this answer

share|cite|improve this answer

edited Feb 26 '11 at 19:45

user242

edited Feb 26 '11 at 19:45

user242

edited Feb 26 '11 at 19:45

user242

answered Feb 26 '11 at 3:10

Number

1

answered Feb 26 '11 at 3:10

Number

1

answered Feb 26 '11 at 3:10

Number

1

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
  – hhh
  Feb 13 '16 at 14:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @hhh Yes, that's what UFD denotes in ring theory.
  – Number
  Jun 4 '17 at 16:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
  – hhh
  Feb 13 '16 at 14:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @hhh Yes, that's what UFD denotes in ring theory.
  – Number
  Jun 4 '17 at 16:18
  

  
  
  
  

What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
– hhh
Feb 13 '16 at 14:08

What does "UFD" mean? en.wikipedia.org/wiki/Unique_factorization_domain ? +1 for the matrix rings' notice
– hhh
Feb 13 '16 at 14:08

1

1

@hhh Yes, that's what UFD denotes in ring theory.
– Number
Jun 4 '17 at 16:18

@hhh Yes, that's what UFD denotes in ring theory.
– Number
Jun 4 '17 at 16:18

add a comment | 

 

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