For a field $F$, what are the elements of $U(F[x])$?

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Let $R$ be a Euclidean domain. An element $u in R$ is said to be a unit if there is a $v in R$ such that $uv = 1$. The set of units in $R$ is denoted by $U(R)$.

Let $R$ be a commutative unital ring. We say that an element $u in R$ is a unit if $u$ has a multiplicative inverse. Prove that the set $U (R) = u R colon ; $u$ text is a unit $ is a group under multiplication.

Certainly $1$ is a unit in $R$, so $U(R)$ contains an identity element. Associativity and commutativity of multiplication is inherited from $R$. Let $u U(R)$. Then there is $u′ in R$ such that $uu′ = 1$. Hence $u′$ is a unit, so $u′ in U(R)$ and $U(R)$ is closed under inverses. Now let $u,v in U(R)$. Therefore there are $u′,v′ in R$ such that $uu′ = 1$ and $vv′ = 1$. Since $uvu′v′ = 1$, it follows that uv is a unit in $R$, so $U(R)$ is closed under multiplication. Hence $U(R)$ is a group.

After proving all of this then my question is: Let $F$ be a field. What are the elements of $U(F[x])$ ?

abstract-algebra field-theory euclidean-domain

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edited Aug 3 at 13:29

quid♦

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asked Aug 3 at 13:09

BeccaBee

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up vote
1
down vote

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Let $R$ be a Euclidean domain. An element $u in R$ is said to be a unit if there is a $v in R$ such that $uv = 1$. The set of units in $R$ is denoted by $U(R)$.

Let $R$ be a commutative unital ring. We say that an element $u in R$ is a unit if $u$ has a multiplicative inverse. Prove that the set $U (R) = u R colon ; $u$ text is a unit $ is a group under multiplication.

Certainly $1$ is a unit in $R$, so $U(R)$ contains an identity element. Associativity and commutativity of multiplication is inherited from $R$. Let $u U(R)$. Then there is $u′ in R$ such that $uu′ = 1$. Hence $u′$ is a unit, so $u′ in U(R)$ and $U(R)$ is closed under inverses. Now let $u,v in U(R)$. Therefore there are $u′,v′ in R$ such that $uu′ = 1$ and $vv′ = 1$. Since $uvu′v′ = 1$, it follows that uv is a unit in $R$, so $U(R)$ is closed under multiplication. Hence $U(R)$ is a group.

After proving all of this then my question is: Let $F$ be a field. What are the elements of $U(F[x])$ ?

abstract-algebra field-theory euclidean-domain

share|cite|improve this question

edited Aug 3 at 13:29

quid♦

36.1k84989

asked Aug 3 at 13:09

BeccaBee

105

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $R$ be a Euclidean domain. An element $u in R$ is said to be a unit if there is a $v in R$ such that $uv = 1$. The set of units in $R$ is denoted by $U(R)$.

Let $R$ be a commutative unital ring. We say that an element $u in R$ is a unit if $u$ has a multiplicative inverse. Prove that the set $U (R) = u R colon ; $u$ text is a unit $ is a group under multiplication.

Certainly $1$ is a unit in $R$, so $U(R)$ contains an identity element. Associativity and commutativity of multiplication is inherited from $R$. Let $u U(R)$. Then there is $u′ in R$ such that $uu′ = 1$. Hence $u′$ is a unit, so $u′ in U(R)$ and $U(R)$ is closed under inverses. Now let $u,v in U(R)$. Therefore there are $u′,v′ in R$ such that $uu′ = 1$ and $vv′ = 1$. Since $uvu′v′ = 1$, it follows that uv is a unit in $R$, so $U(R)$ is closed under multiplication. Hence $U(R)$ is a group.

After proving all of this then my question is: Let $F$ be a field. What are the elements of $U(F[x])$ ?

abstract-algebra field-theory euclidean-domain

share|cite|improve this question

edited Aug 3 at 13:29

quid♦

36.1k84989

asked Aug 3 at 13:09

BeccaBee

105

Let $R$ be a Euclidean domain. An element $u in R$ is said to be a unit if there is a $v in R$ such that $uv = 1$. The set of units in $R$ is denoted by $U(R)$.

Let $R$ be a commutative unital ring. We say that an element $u in R$ is a unit if $u$ has a multiplicative inverse. Prove that the set $U (R) = u R colon ; $u$ text is a unit $ is a group under multiplication.

Certainly $1$ is a unit in $R$, so $U(R)$ contains an identity element. Associativity and commutativity of multiplication is inherited from $R$. Let $u U(R)$. Then there is $u′ in R$ such that $uu′ = 1$. Hence $u′$ is a unit, so $u′ in U(R)$ and $U(R)$ is closed under inverses. Now let $u,v in U(R)$. Therefore there are $u′,v′ in R$ such that $uu′ = 1$ and $vv′ = 1$. Since $uvu′v′ = 1$, it follows that uv is a unit in $R$, so $U(R)$ is closed under multiplication. Hence $U(R)$ is a group.

After proving all of this then my question is: Let $F$ be a field. What are the elements of $U(F[x])$ ?

abstract-algebra field-theory euclidean-domain

share|cite|improve this question

edited Aug 3 at 13:29

quid♦

36.1k84989

asked Aug 3 at 13:09

BeccaBee

105

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 13:29

quid♦

36.1k84989

edited Aug 3 at 13:29

quid♦

36.1k84989

edited Aug 3 at 13:29

quid♦

36.1k84989

36.1k84989

asked Aug 3 at 13:09

BeccaBee

105

asked Aug 3 at 13:09

BeccaBee

105

asked Aug 3 at 13:09

BeccaBee

105

105

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F setminus 0$.

It remains to consider which elements of $F setminus 0$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F setminus 0$ has a multiplicative inverse.

Thus, $U(F[X]) = F setminus 0$.

This argument readily generalizes to show that for $R$ an integral domain
$U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

share|cite|improve this answer

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I was over complicating things and thinking that it was too trivial.
  – BeccaBee
  Aug 3 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Chances are the question is already answered somewhere on the site.
  – quid♦
  Aug 3 at 13:31
  

  
  
  
  

add a comment | 

up vote
1
down vote

We claim that
$$U(F[x])=U(F)=F-0.$$
Of course, $U(F) subseteq U(F[x])$. Conversely, if $p(x) in U(F[x])$, then by definition there exists $q(x) in F[x]$ such that
$$p(x)q(x)=1.$$
In particular:
$$deg(p(x)q(x))=deg(1)$$
$$deg(p(x))+deg(q(x))=0,$$
which forces $deg(p(x))=deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) in F subset F[x]$. Since $pq=1$, it follows that $p in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then
$$U(R[x])=U(R).$$

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It seems I arrived late ;)
  – Don
  Aug 3 at 13:40
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F setminus 0$.

It remains to consider which elements of $F setminus 0$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F setminus 0$ has a multiplicative inverse.

Thus, $U(F[X]) = F setminus 0$.

This argument readily generalizes to show that for $R$ an integral domain
$U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

share|cite|improve this answer

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I was over complicating things and thinking that it was too trivial.
  – BeccaBee
  Aug 3 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Chances are the question is already answered somewhere on the site.
  – quid♦
  Aug 3 at 13:31
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F setminus 0$.

It remains to consider which elements of $F setminus 0$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F setminus 0$ has a multiplicative inverse.

Thus, $U(F[X]) = F setminus 0$.

This argument readily generalizes to show that for $R$ an integral domain
$U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

share|cite|improve this answer

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I was over complicating things and thinking that it was too trivial.
  – BeccaBee
  Aug 3 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Chances are the question is already answered somewhere on the site.
  – quid♦
  Aug 3 at 13:31
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F setminus 0$.

It remains to consider which elements of $F setminus 0$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F setminus 0$ has a multiplicative inverse.

Thus, $U(F[X]) = F setminus 0$.

This argument readily generalizes to show that for $R$ an integral domain
$U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

share|cite|improve this answer

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F setminus 0$.

It remains to consider which elements of $F setminus 0$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F setminus 0$ has a multiplicative inverse.

Thus, $U(F[X]) = F setminus 0$.

This argument readily generalizes to show that for $R$ an integral domain
$U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

share|cite|improve this answer

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 14:23

edited Aug 3 at 14:23

edited Aug 3 at 14:23

answered Aug 3 at 13:24

quid♦

36.1k84989

answered Aug 3 at 13:24

quid♦

36.1k84989

answered Aug 3 at 13:24

quid♦

36.1k84989

36.1k84989

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I was over complicating things and thinking that it was too trivial.
  – BeccaBee
  Aug 3 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Chances are the question is already answered somewhere on the site.
  – quid♦
  Aug 3 at 13:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I was over complicating things and thinking that it was too trivial.
  – BeccaBee
  Aug 3 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Chances are the question is already answered somewhere on the site.
  – quid♦
  Aug 3 at 13:31
  

  
  
  
  

Thank you. I was over complicating things and thinking that it was too trivial.
– BeccaBee
Aug 3 at 13:29

Thank you. I was over complicating things and thinking that it was too trivial.
– BeccaBee
Aug 3 at 13:29

You are welcome. Chances are the question is already answered somewhere on the site.
– quid♦
Aug 3 at 13:31

You are welcome. Chances are the question is already answered somewhere on the site.
– quid♦
Aug 3 at 13:31

add a comment | 

up vote
1
down vote

We claim that
$$U(F[x])=U(F)=F-0.$$
Of course, $U(F) subseteq U(F[x])$. Conversely, if $p(x) in U(F[x])$, then by definition there exists $q(x) in F[x]$ such that
$$p(x)q(x)=1.$$
In particular:
$$deg(p(x)q(x))=deg(1)$$
$$deg(p(x))+deg(q(x))=0,$$
which forces $deg(p(x))=deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) in F subset F[x]$. Since $pq=1$, it follows that $p in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then
$$U(R[x])=U(R).$$

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It seems I arrived late ;)
  – Don
  Aug 3 at 13:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

We claim that
$$U(F[x])=U(F)=F-0.$$
Of course, $U(F) subseteq U(F[x])$. Conversely, if $p(x) in U(F[x])$, then by definition there exists $q(x) in F[x]$ such that
$$p(x)q(x)=1.$$
In particular:
$$deg(p(x)q(x))=deg(1)$$
$$deg(p(x))+deg(q(x))=0,$$
which forces $deg(p(x))=deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) in F subset F[x]$. Since $pq=1$, it follows that $p in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then
$$U(R[x])=U(R).$$

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It seems I arrived late ;)
  – Don
  Aug 3 at 13:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

We claim that
$$U(F[x])=U(F)=F-0.$$
Of course, $U(F) subseteq U(F[x])$. Conversely, if $p(x) in U(F[x])$, then by definition there exists $q(x) in F[x]$ such that
$$p(x)q(x)=1.$$
In particular:
$$deg(p(x)q(x))=deg(1)$$
$$deg(p(x))+deg(q(x))=0,$$
which forces $deg(p(x))=deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) in F subset F[x]$. Since $pq=1$, it follows that $p in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then
$$U(R[x])=U(R).$$

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

We claim that
$$U(F[x])=U(F)=F-0.$$
Of course, $U(F) subseteq U(F[x])$. Conversely, if $p(x) in U(F[x])$, then by definition there exists $q(x) in F[x]$ such that
$$p(x)q(x)=1.$$
In particular:
$$deg(p(x)q(x))=deg(1)$$
$$deg(p(x))+deg(q(x))=0,$$
which forces $deg(p(x))=deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) in F subset F[x]$. Since $pq=1$, it follows that $p in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then
$$U(R[x])=U(R).$$

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 13:37

Don

1263

answered Aug 3 at 13:37

Don

1263

answered Aug 3 at 13:37

Don

1263

1263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It seems I arrived late ;)
  – Don
  Aug 3 at 13:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It seems I arrived late ;)
  – Don
  Aug 3 at 13:40
  

  
  
  
  

It seems I arrived late ;)
– Don
Aug 3 at 13:40

It seems I arrived late ;)
– Don
Aug 3 at 13:40

add a comment | 

 

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