Mixing
Quadratic integers
Clash Royale CLAN TAG#URR8PPP
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Let consider the integer of $mathbbQ(sqrtd)$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $mathcalO=mathbbZ+mathbbZsqrtd$. If $d=4k+1$, it is $mathcalO=mathbbZ+mathbbZfrac1+sqrtd2$.
If $mathcalP$ is a non-zero prime ideal of $mathcalO$, then we have $mathcalPcapmathbbZ=pmathbbZ$ for some prime number $p$. What I want to prove is $mathcalO/mathcalP$ is a field, actually an at-most-two-degree extension of $mathbbF_p$.
If we pick $[x]neq[0]inmathcalO/mathcalP$, we need to show $xnotin mathcalP$, there exists $yin mathcalO$, such that $[yx]=1 text mod p$.
Further, if we consider the norm of $x=a+bsqrtd$, how can we deduce $n(x)=barxx$ and $p$ are coprime? If we can deduce this property then $y:=c(a-bsqrtd)$ can be an inverse of $x$.
abstract-algebra number-theory algebraic-number-theory
asked Jul 27 at 14:15
Xavier Yang
440314
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3
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Let consider the integer of $mathbbQ(sqrtd)$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $mathcalO=mathbbZ+mathbbZsqrtd$. If $d=4k+1$, it is $mathcalO=mathbbZ+mathbbZfrac1+sqrtd2$.
If $mathcalP$ is a non-zero prime ideal of $mathcalO$, then we have $mathcalPcapmathbbZ=pmathbbZ$ for some prime number $p$. What I want to prove is $mathcalO/mathcalP$ is a field, actually an at-most-two-degree extension of $mathbbF_p$.
If we pick $[x]neq[0]inmathcalO/mathcalP$, we need to show $xnotin mathcalP$, there exists $yin mathcalO$, such that $[yx]=1 text mod p$.
Further, if we consider the norm of $x=a+bsqrtd$, how can we deduce $n(x)=barxx$ and $p$ are coprime? If we can deduce this property then $y:=c(a-bsqrtd)$ can be an inverse of $x$.
abstract-algebra number-theory algebraic-number-theory
asked Jul 27 at 14:15
Xavier Yang
440314
So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let consider the integer of $mathbbQ(sqrtd)$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $mathcalO=mathbbZ+mathbbZsqrtd$. If $d=4k+1$, it is $mathcalO=mathbbZ+mathbbZfrac1+sqrtd2$.
If $mathcalP$ is a non-zero prime ideal of $mathcalO$, then we have $mathcalPcapmathbbZ=pmathbbZ$ for some prime number $p$. What I want to prove is $mathcalO/mathcalP$ is a field, actually an at-most-two-degree extension of $mathbbF_p$.
If we pick $[x]neq[0]inmathcalO/mathcalP$, we need to show $xnotin mathcalP$, there exists $yin mathcalO$, such that $[yx]=1 text mod p$.
Further, if we consider the norm of $x=a+bsqrtd$, how can we deduce $n(x)=barxx$ and $p$ are coprime? If we can deduce this property then $y:=c(a-bsqrtd)$ can be an inverse of $x$.
abstract-algebra number-theory algebraic-number-theory
asked Jul 27 at 14:15
Xavier Yang
440314
Let consider the integer of $mathbbQ(sqrtd)$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $mathcalO=mathbbZ+mathbbZsqrtd$. If $d=4k+1$, it is $mathcalO=mathbbZ+mathbbZfrac1+sqrtd2$.
If $mathcalP$ is a non-zero prime ideal of $mathcalO$, then we have $mathcalPcapmathbbZ=pmathbbZ$ for some prime number $p$. What I want to prove is $mathcalO/mathcalP$ is a field, actually an at-most-two-degree extension of $mathbbF_p$.
If we pick $[x]neq[0]inmathcalO/mathcalP$, we need to show $xnotin mathcalP$, there exists $yin mathcalO$, such that $[yx]=1 text mod p$.
Further, if we consider the norm of $x=a+bsqrtd$, how can we deduce $n(x)=barxx$ and $p$ are coprime? If we can deduce this property then $y:=c(a-bsqrtd)$ can be an inverse of $x$.
abstract-algebra number-theory algebraic-number-theory
asked Jul 27 at 14:15
Xavier Yang
440314
asked Jul 27 at 14:15
Xavier Yang
440314
asked Jul 27 at 14:15
Xavier Yang
440314
asked Jul 27 at 14:15
Xavier Yang
440314
440314
So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34
add a comment |Â
So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34
So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34
add a comment |Â
2 Answers
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Let $a$ be the algebraic integer generating $K=Bbb Q(sqrt d)$, so $a$ is either $sqrt d$, or $(1+sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $Bbb Z[X]$. Then
$$
beginaligned
K &= Bbb Q(a) ,\
mathcal O_K &= Bbb Z[a]=Bbb Z[X]/f ,\
mathcal O_K/mathcal P &= Bbb Z/p=Bbb F_p ,\
&text for a suitable unique prime $p$, for $mathcal O_K/mathcal P$
is a field, for $mathcal P$ is prime,
\
mathcal O_K/p
&= Bbb Z[a]/p\
&= (Bbb Z[X]/f)/p\
&= Bbb Z[X]/(f,p)\
&= (Bbb Z[X]/p)/f\
&= (Bbb Z/p)[X]/f\
&= Bbb F_p[X]/f .
endaligned
$$
Now i try to answer the posted question in a discussion. Since $mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,nin Bbb Z$, as explicit as possible.
Examples may make the situation transparent. We work in $Bbb Q(sqrt 2018)$, so $a= sqrt 2018$. Some primes in $Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:
sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....: print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....: % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:
Is p = 2 prime? False :: Is 2018 a square mod p? True
Is p = 3 prime? True :: Is 2018 a square mod p? False
Is p = 5 prime? True :: Is 2018 a square mod p? False
Is p = 7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime? True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime? True :: Is 2018 a square mod p? False
Is p = 29 prime? True :: Is 2018 a square mod p? False
Is p = 31 prime? True :: Is 2018 a square mod p? False
Is p = 37 prime? True :: Is 2018 a square mod p? False
The pattern is transparent.
(We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $mathcal O_K$, and conversely.
Consider for instance first $p=5$. Then with the above, it is a prime, since $mathcal O_K/5cong Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $Bbb O_K$. How to get the inverse of an element $xi$ in this field? As the OP suggests, we build the conjugate $barxi$ and have to show that we do not get the product zero in $xi,barxi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.
Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $mathcal O_K/7cong Bbb F_7[X]/(X^2-2018)=Bbb F_7[X]/(X^2-9)cong Bbb F_7[X]/(X-3)times Bbb F_7[X]/(X+3)$,
and we can also factor:sage: K.<a> = QuadraticField(2018)
sage: K(7).factor()
(-1) * (a - 45) * (a + 45)
so $7$ is not irreducible. The primes over $7$ are principal ideals,
generated by $45pm a$ each.
How to find an inverse in the case $mathcal P=(45-a)$? (The other case is similar, by conjugation.)
We choose a $xi=m+na$ which is not zero mod $(45-a)$. Then modulo $mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.
Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,
sage: K.ideal(11).factor()
(Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
and both ideals over $11$ in $mathcal O_K$ contain an element of the shape $apm 4$, again we replace $a$ in an element $m+na$ modulo $mathcal P$ by the corresponding $pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.
answered Jul 29 at 18:57
dan_fulea
4,1171211
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Here is a direct proof. As you noticed, for any non null prime ideal $mathcal P$ of the ring of integers $mathcal O$, we have $mathcal O cap mathbf Z=pmathbf Z $ for a certain rational prime $p$. Considering $mathbf Z/pmathbf Z$ as a subring of $mathcal O/mathcal P$ and using $mathcal O=mathbf Z[alpha]$, we see that $mathcal O/mathcal P$ is an $mathbf F_p$-vector space of dimension at most 2, hence is finite. But $mathcal O/mathcal P$ is a domain since $mathcal P$ is prime, and a finite domain is necessarily a field.
answered Jul 28 at 13:02
nguyen quang do
7,4371621
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $a$ be the algebraic integer generating $K=Bbb Q(sqrt d)$, so $a$ is either $sqrt d$, or $(1+sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $Bbb Z[X]$. Then
$$
beginaligned
K &= Bbb Q(a) ,\
mathcal O_K &= Bbb Z[a]=Bbb Z[X]/f ,\
mathcal O_K/mathcal P &= Bbb Z/p=Bbb F_p ,\
&text for a suitable unique prime $p$, for $mathcal O_K/mathcal P$
is a field, for $mathcal P$ is prime,
\
mathcal O_K/p
&= Bbb Z[a]/p\
&= (Bbb Z[X]/f)/p\
&= Bbb Z[X]/(f,p)\
&= (Bbb Z[X]/p)/f\
&= (Bbb Z/p)[X]/f\
&= Bbb F_p[X]/f .
endaligned
$$
Now i try to answer the posted question in a discussion. Since $mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,nin Bbb Z$, as explicit as possible.
Examples may make the situation transparent. We work in $Bbb Q(sqrt 2018)$, so $a= sqrt 2018$. Some primes in $Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:
sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....: print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....: % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:
Is p = 2 prime? False :: Is 2018 a square mod p? True
Is p = 3 prime? True :: Is 2018 a square mod p? False
Is p = 5 prime? True :: Is 2018 a square mod p? False
Is p = 7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime? True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime? True :: Is 2018 a square mod p? False
Is p = 29 prime? True :: Is 2018 a square mod p? False
Is p = 31 prime? True :: Is 2018 a square mod p? False
Is p = 37 prime? True :: Is 2018 a square mod p? False
The pattern is transparent.
(We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $mathcal O_K$, and conversely.
Consider for instance first $p=5$. Then with the above, it is a prime, since $mathcal O_K/5cong Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $Bbb O_K$. How to get the inverse of an element $xi$ in this field? As the OP suggests, we build the conjugate $barxi$ and have to show that we do not get the product zero in $xi,barxi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.
Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $mathcal O_K/7cong Bbb F_7[X]/(X^2-2018)=Bbb F_7[X]/(X^2-9)cong Bbb F_7[X]/(X-3)times Bbb F_7[X]/(X+3)$,
and we can also factor:sage: K.<a> = QuadraticField(2018)
sage: K(7).factor()
(-1) * (a - 45) * (a + 45)
so $7$ is not irreducible. The primes over $7$ are principal ideals,
generated by $45pm a$ each.
How to find an inverse in the case $mathcal P=(45-a)$? (The other case is similar, by conjugation.)
We choose a $xi=m+na$ which is not zero mod $(45-a)$. Then modulo $mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.
Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,
sage: K.ideal(11).factor()
(Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
and both ideals over $11$ in $mathcal O_K$ contain an element of the shape $apm 4$, again we replace $a$ in an element $m+na$ modulo $mathcal P$ by the corresponding $pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.
answered Jul 29 at 18:57
dan_fulea
4,1171211
add a comment |Â
up vote
3
down vote
Let $a$ be the algebraic integer generating $K=Bbb Q(sqrt d)$, so $a$ is either $sqrt d$, or $(1+sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $Bbb Z[X]$. Then
$$
beginaligned
K &= Bbb Q(a) ,\
mathcal O_K &= Bbb Z[a]=Bbb Z[X]/f ,\
mathcal O_K/mathcal P &= Bbb Z/p=Bbb F_p ,\
&text for a suitable unique prime $p$, for $mathcal O_K/mathcal P$
is a field, for $mathcal P$ is prime,
\
mathcal O_K/p
&= Bbb Z[a]/p\
&= (Bbb Z[X]/f)/p\
&= Bbb Z[X]/(f,p)\
&= (Bbb Z[X]/p)/f\
&= (Bbb Z/p)[X]/f\
&= Bbb F_p[X]/f .
endaligned
$$
Now i try to answer the posted question in a discussion. Since $mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,nin Bbb Z$, as explicit as possible.
Examples may make the situation transparent. We work in $Bbb Q(sqrt 2018)$, so $a= sqrt 2018$. Some primes in $Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:
sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....: print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....: % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:
Is p = 2 prime? False :: Is 2018 a square mod p? True
Is p = 3 prime? True :: Is 2018 a square mod p? False
Is p = 5 prime? True :: Is 2018 a square mod p? False
Is p = 7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime? True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime? True :: Is 2018 a square mod p? False
Is p = 29 prime? True :: Is 2018 a square mod p? False
Is p = 31 prime? True :: Is 2018 a square mod p? False
Is p = 37 prime? True :: Is 2018 a square mod p? False
The pattern is transparent.
(We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $mathcal O_K$, and conversely.
Consider for instance first $p=5$. Then with the above, it is a prime, since $mathcal O_K/5cong Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $Bbb O_K$. How to get the inverse of an element $xi$ in this field? As the OP suggests, we build the conjugate $barxi$ and have to show that we do not get the product zero in $xi,barxi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.
Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $mathcal O_K/7cong Bbb F_7[X]/(X^2-2018)=Bbb F_7[X]/(X^2-9)cong Bbb F_7[X]/(X-3)times Bbb F_7[X]/(X+3)$,
and we can also factor:sage: K.<a> = QuadraticField(2018)
sage: K(7).factor()
(-1) * (a - 45) * (a + 45)
so $7$ is not irreducible. The primes over $7$ are principal ideals,
generated by $45pm a$ each.
How to find an inverse in the case $mathcal P=(45-a)$? (The other case is similar, by conjugation.)
We choose a $xi=m+na$ which is not zero mod $(45-a)$. Then modulo $mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.
Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,
sage: K.ideal(11).factor()
(Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
and both ideals over $11$ in $mathcal O_K$ contain an element of the shape $apm 4$, again we replace $a$ in an element $m+na$ modulo $mathcal P$ by the corresponding $pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.
answered Jul 29 at 18:57
dan_fulea
4,1171211
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $a$ be the algebraic integer generating $K=Bbb Q(sqrt d)$, so $a$ is either $sqrt d$, or $(1+sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $Bbb Z[X]$. Then
$$
beginaligned
K &= Bbb Q(a) ,\
mathcal O_K &= Bbb Z[a]=Bbb Z[X]/f ,\
mathcal O_K/mathcal P &= Bbb Z/p=Bbb F_p ,\
&text for a suitable unique prime $p$, for $mathcal O_K/mathcal P$
is a field, for $mathcal P$ is prime,
\
mathcal O_K/p
&= Bbb Z[a]/p\
&= (Bbb Z[X]/f)/p\
&= Bbb Z[X]/(f,p)\
&= (Bbb Z[X]/p)/f\
&= (Bbb Z/p)[X]/f\
&= Bbb F_p[X]/f .
endaligned
$$
Now i try to answer the posted question in a discussion. Since $mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,nin Bbb Z$, as explicit as possible.
Examples may make the situation transparent. We work in $Bbb Q(sqrt 2018)$, so $a= sqrt 2018$. Some primes in $Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:
sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....: print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....: % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:
Is p = 2 prime? False :: Is 2018 a square mod p? True
Is p = 3 prime? True :: Is 2018 a square mod p? False
Is p = 5 prime? True :: Is 2018 a square mod p? False
Is p = 7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime? True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime? True :: Is 2018 a square mod p? False
Is p = 29 prime? True :: Is 2018 a square mod p? False
Is p = 31 prime? True :: Is 2018 a square mod p? False
Is p = 37 prime? True :: Is 2018 a square mod p? False
The pattern is transparent.
(We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $mathcal O_K$, and conversely.
Consider for instance first $p=5$. Then with the above, it is a prime, since $mathcal O_K/5cong Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $Bbb O_K$. How to get the inverse of an element $xi$ in this field? As the OP suggests, we build the conjugate $barxi$ and have to show that we do not get the product zero in $xi,barxi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.
Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $mathcal O_K/7cong Bbb F_7[X]/(X^2-2018)=Bbb F_7[X]/(X^2-9)cong Bbb F_7[X]/(X-3)times Bbb F_7[X]/(X+3)$,
and we can also factor:sage: K.<a> = QuadraticField(2018)
sage: K(7).factor()
(-1) * (a - 45) * (a + 45)
so $7$ is not irreducible. The primes over $7$ are principal ideals,
generated by $45pm a$ each.
How to find an inverse in the case $mathcal P=(45-a)$? (The other case is similar, by conjugation.)
We choose a $xi=m+na$ which is not zero mod $(45-a)$. Then modulo $mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.
Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,
sage: K.ideal(11).factor()
(Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
and both ideals over $11$ in $mathcal O_K$ contain an element of the shape $apm 4$, again we replace $a$ in an element $m+na$ modulo $mathcal P$ by the corresponding $pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.
answered Jul 29 at 18:57
dan_fulea
4,1171211
Let $a$ be the algebraic integer generating $K=Bbb Q(sqrt d)$, so $a$ is either $sqrt d$, or $(1+sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $Bbb Z[X]$. Then
$$
beginaligned
K &= Bbb Q(a) ,\
mathcal O_K &= Bbb Z[a]=Bbb Z[X]/f ,\
mathcal O_K/mathcal P &= Bbb Z/p=Bbb F_p ,\
&text for a suitable unique prime $p$, for $mathcal O_K/mathcal P$
is a field, for $mathcal P$ is prime,
\
mathcal O_K/p
&= Bbb Z[a]/p\
&= (Bbb Z[X]/f)/p\
&= Bbb Z[X]/(f,p)\
&= (Bbb Z[X]/p)/f\
&= (Bbb Z/p)[X]/f\
&= Bbb F_p[X]/f .
endaligned
$$
Now i try to answer the posted question in a discussion. Since $mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,nin Bbb Z$, as explicit as possible.
Examples may make the situation transparent. We work in $Bbb Q(sqrt 2018)$, so $a= sqrt 2018$. Some primes in $Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:
sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....: print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....: % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:
Is p = 2 prime? False :: Is 2018 a square mod p? True
Is p = 3 prime? True :: Is 2018 a square mod p? False
Is p = 5 prime? True :: Is 2018 a square mod p? False
Is p = 7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime? True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime? True :: Is 2018 a square mod p? False
Is p = 29 prime? True :: Is 2018 a square mod p? False
Is p = 31 prime? True :: Is 2018 a square mod p? False
Is p = 37 prime? True :: Is 2018 a square mod p? False
The pattern is transparent.
(We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $mathcal O_K$, and conversely.
Consider for instance first $p=5$. Then with the above, it is a prime, since $mathcal O_K/5cong Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $Bbb O_K$. How to get the inverse of an element $xi$ in this field? As the OP suggests, we build the conjugate $barxi$ and have to show that we do not get the product zero in $xi,barxi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.
Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $mathcal O_K/7cong Bbb F_7[X]/(X^2-2018)=Bbb F_7[X]/(X^2-9)cong Bbb F_7[X]/(X-3)times Bbb F_7[X]/(X+3)$,
and we can also factor:sage: K.<a> = QuadraticField(2018)
sage: K(7).factor()
(-1) * (a - 45) * (a + 45)
so $7$ is not irreducible. The primes over $7$ are principal ideals,
generated by $45pm a$ each.
How to find an inverse in the case $mathcal P=(45-a)$? (The other case is similar, by conjugation.)
We choose a $xi=m+na$ which is not zero mod $(45-a)$. Then modulo $mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.
Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,
sage: K.ideal(11).factor()
(Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
and both ideals over $11$ in $mathcal O_K$ contain an element of the shape $apm 4$, again we replace $a$ in an element $m+na$ modulo $mathcal P$ by the corresponding $pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.
answered Jul 29 at 18:57
dan_fulea
4,1171211
answered Jul 29 at 18:57
dan_fulea
4,1171211
answered Jul 29 at 18:57
dan_fulea
4,1171211
answered Jul 29 at 18:57
dan_fulea
4,1171211
4,1171211
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Here is a direct proof. As you noticed, for any non null prime ideal $mathcal P$ of the ring of integers $mathcal O$, we have $mathcal O cap mathbf Z=pmathbf Z $ for a certain rational prime $p$. Considering $mathbf Z/pmathbf Z$ as a subring of $mathcal O/mathcal P$ and using $mathcal O=mathbf Z[alpha]$, we see that $mathcal O/mathcal P$ is an $mathbf F_p$-vector space of dimension at most 2, hence is finite. But $mathcal O/mathcal P$ is a domain since $mathcal P$ is prime, and a finite domain is necessarily a field.
answered Jul 28 at 13:02
nguyen quang do
7,4371621
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up vote
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Here is a direct proof. As you noticed, for any non null prime ideal $mathcal P$ of the ring of integers $mathcal O$, we have $mathcal O cap mathbf Z=pmathbf Z $ for a certain rational prime $p$. Considering $mathbf Z/pmathbf Z$ as a subring of $mathcal O/mathcal P$ and using $mathcal O=mathbf Z[alpha]$, we see that $mathcal O/mathcal P$ is an $mathbf F_p$-vector space of dimension at most 2, hence is finite. But $mathcal O/mathcal P$ is a domain since $mathcal P$ is prime, and a finite domain is necessarily a field.
answered Jul 28 at 13:02
nguyen quang do
7,4371621
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a direct proof. As you noticed, for any non null prime ideal $mathcal P$ of the ring of integers $mathcal O$, we have $mathcal O cap mathbf Z=pmathbf Z $ for a certain rational prime $p$. Considering $mathbf Z/pmathbf Z$ as a subring of $mathcal O/mathcal P$ and using $mathcal O=mathbf Z[alpha]$, we see that $mathcal O/mathcal P$ is an $mathbf F_p$-vector space of dimension at most 2, hence is finite. But $mathcal O/mathcal P$ is a domain since $mathcal P$ is prime, and a finite domain is necessarily a field.
answered Jul 28 at 13:02
nguyen quang do
7,4371621
Here is a direct proof. As you noticed, for any non null prime ideal $mathcal P$ of the ring of integers $mathcal O$, we have $mathcal O cap mathbf Z=pmathbf Z $ for a certain rational prime $p$. Considering $mathbf Z/pmathbf Z$ as a subring of $mathcal O/mathcal P$ and using $mathcal O=mathbf Z[alpha]$, we see that $mathcal O/mathcal P$ is an $mathbf F_p$-vector space of dimension at most 2, hence is finite. But $mathcal O/mathcal P$ is a domain since $mathcal P$ is prime, and a finite domain is necessarily a field.
answered Jul 28 at 13:02
nguyen quang do
7,4371621
answered Jul 28 at 13:02
nguyen quang do
7,4371621
answered Jul 28 at 13:02
nguyen quang do
7,4371621
answered Jul 28 at 13:02
nguyen quang do
7,4371621
7,4371621
add a comment |Â
add a comment |Â
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So you know $Z/pto O/P$ is integral. And $O/P$ is a domain and $Z/p$ is a field. Then integrality says $O/P$ is a field.
– user45765
Jul 27 at 21:56
The ring of integers of a number field is a Dedekind ring, and in a Dedekind ring, every non zero prime ideal is maximal. But I suppose that you wanted a direct proof in the case of a quadratic field ?
– nguyen quang do
Jul 28 at 12:34