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Let $R$ be a domain. Prove that if a polynomial in $R[x]$ is a unit, then it is a nonzero constant (the converse is true if $R$ is a field)

Is this a property of an integral domain that is not a field?Prove that $D[x]$ is an integral domain if $D$ is one.If every nonzero element of $R$ is either a unit or a zero divisor then $R$ contains only finitely many ideals?Prove that if R[x] is a PID, then R is a fieldLet $R$ be an integral domain which is not a field. Then can $R[x]$ have a maximal ideal generated by a non-constant polynomial?Prove that an integer monic polynomial with nonzero constant term and exactly one root outside the open unit disk is irreducible over $mathbbQ$If $D[x]$ is a Prüfer domain, then $D$ is a fieldLet $K$ be a Field and $Ain M_2(mathbbK)$, $A not = 0.$ Show that $A$ is a unit iff $A$ is a left non-zero divisor in $M_2(mathbbK).$If $R$ is an integral domain then $R[x]$ forms an integral domainIs it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?

0

$begingroup$

Problem: Let $R$ be a domain. Prove that if a polynomial in $R[x]$ is a unit, then it is a nonzero constant (the converse is true if $R$
is a field).

My attempt: We proof that by contradiction. Suppose $u in R$ be a unit of $R$ but $u$ is not a nonzero constant, then we have $deg (u) > 0$.

$forall f in R$, $deg (uf) leq deg (u) + deg (f)$.

On the other hand, $u$ is a unit of $R$, so $deg (uf) = deg (f)$.

Associate with the inequality above, we see that the equality hold if and only if $deg (u) = 0$. So we can conclude that $u$ is a nonzero constant.

Is my proof correct??? Thanks all!!!

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

asked Mar 12 at 8:17

MinhMinh

31119

$endgroup$

add a comment | 

0

$begingroup$

Problem: Let $R$ be a domain. Prove that if a polynomial in $R[x]$ is a unit, then it is a nonzero constant (the converse is true if $R$
is a field).

My attempt: We proof that by contradiction. Suppose $u in R$ be a unit of $R$ but $u$ is not a nonzero constant, then we have $deg (u) > 0$.

$forall f in R$, $deg (uf) leq deg (u) + deg (f)$.

On the other hand, $u$ is a unit of $R$, so $deg (uf) = deg (f)$.

Associate with the inequality above, we see that the equality hold if and only if $deg (u) = 0$. So we can conclude that $u$ is a nonzero constant.

Is my proof correct??? Thanks all!!!

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

asked Mar 12 at 8:17

MinhMinh

31119

$endgroup$

add a comment | 

$begingroup$

Problem: Let $R$ be a domain. Prove that if a polynomial in $R[x]$ is a unit, then it is a nonzero constant (the converse is true if $R$
is a field).

My attempt: We proof that by contradiction. Suppose $u in R$ be a unit of $R$ but $u$ is not a nonzero constant, then we have $deg (u) > 0$.

$forall f in R$, $deg (uf) leq deg (u) + deg (f)$.

On the other hand, $u$ is a unit of $R$, so $deg (uf) = deg (f)$.

Associate with the inequality above, we see that the equality hold if and only if $deg (u) = 0$. So we can conclude that $u$ is a nonzero constant.

Is my proof correct??? Thanks all!!!

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

asked Mar 12 at 8:17

MinhMinh

31119

$endgroup$

share|cite|improve this question

asked Mar 12 at 8:17

MinhMinh

31119

share|cite|improve this question

asked Mar 12 at 8:17

MinhMinh

31119

asked Mar 12 at 8:17

MinhMinh

31119

asked Mar 12 at 8:17

MinhMinh

31119

1 Answer
1

active

oldest

votes

2

$begingroup$

Well, a proof by contradition is not necessary. Suppose $f$ is a unit with inverse $g$. Then $fg=1$.
Using degrees, we obtain
$$0 = rm deg(1) = rm deg(fg) = rm deg(f) + rm deg(g).$$
The last equality holds since $R$ has no zero divisors.
As the degree is a nonnegative function, it follows that both, $f$ and $g$ have degree zero and so are constants (elements of $R$).

share|cite|improve this answer

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the unit has inverse?
  $endgroup$
  – Minh
  Mar 12 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed. An element $fin R$ is a unit if there is an element $gin R$ with $fg=1=fg$. The inverse $g$ is uniquely determined.
  $endgroup$
  – Wuestenfux
  Mar 12 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my opinion, an element $f in R$ is a unit if $forall g in R$, we have $fg = 1$.
  $endgroup$
  – Minh
  Mar 12 at 8:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that, we are considering $R$ as a domain, which is a commutative ring has unit $1 neq 0$ and has the zero-product property. So, the inverse of unit here, i'm not sure it has.
  $endgroup$
  – Minh
  Mar 12 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Minh Did you mean $forall gin R$ we have $fg = g$? That's the definition of $1$, sure. But (perhaps unfortunately) that's not the behaviour we want to capture with the word "unit". An element of a ring is a unit if it's invertible, not necessarily the multiplicative identity.
  $endgroup$
  – Arthur
  Mar 12 at 8:44
  
  

  
  
  
  

 | 
show 1 more comment

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2

$begingroup$

Well, a proof by contradition is not necessary. Suppose $f$ is a unit with inverse $g$. Then $fg=1$.
Using degrees, we obtain
$$0 = rm deg(1) = rm deg(fg) = rm deg(f) + rm deg(g).$$
The last equality holds since $R$ has no zero divisors.
As the degree is a nonnegative function, it follows that both, $f$ and $g$ have degree zero and so are constants (elements of $R$).

share|cite|improve this answer

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the unit has inverse?
  $endgroup$
  – Minh
  Mar 12 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed. An element $fin R$ is a unit if there is an element $gin R$ with $fg=1=fg$. The inverse $g$ is uniquely determined.
  $endgroup$
  – Wuestenfux
  Mar 12 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my opinion, an element $f in R$ is a unit if $forall g in R$, we have $fg = 1$.
  $endgroup$
  – Minh
  Mar 12 at 8:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that, we are considering $R$ as a domain, which is a commutative ring has unit $1 neq 0$ and has the zero-product property. So, the inverse of unit here, i'm not sure it has.
  $endgroup$
  – Minh
  Mar 12 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Minh Did you mean $forall gin R$ we have $fg = g$? That's the definition of $1$, sure. But (perhaps unfortunately) that's not the behaviour we want to capture with the word "unit". An element of a ring is a unit if it's invertible, not necessarily the multiplicative identity.
  $endgroup$
  – Arthur
  Mar 12 at 8:44
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Well, a proof by contradition is not necessary. Suppose $f$ is a unit with inverse $g$. Then $fg=1$.
Using degrees, we obtain
$$0 = rm deg(1) = rm deg(fg) = rm deg(f) + rm deg(g).$$
The last equality holds since $R$ has no zero divisors.
As the degree is a nonnegative function, it follows that both, $f$ and $g$ have degree zero and so are constants (elements of $R$).

share|cite|improve this answer

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the unit has inverse?
  $endgroup$
  – Minh
  Mar 12 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed. An element $fin R$ is a unit if there is an element $gin R$ with $fg=1=fg$. The inverse $g$ is uniquely determined.
  $endgroup$
  – Wuestenfux
  Mar 12 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my opinion, an element $f in R$ is a unit if $forall g in R$, we have $fg = 1$.
  $endgroup$
  – Minh
  Mar 12 at 8:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that, we are considering $R$ as a domain, which is a commutative ring has unit $1 neq 0$ and has the zero-product property. So, the inverse of unit here, i'm not sure it has.
  $endgroup$
  – Minh
  Mar 12 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Minh Did you mean $forall gin R$ we have $fg = g$? That's the definition of $1$, sure. But (perhaps unfortunately) that's not the behaviour we want to capture with the word "unit". An element of a ring is a unit if it's invertible, not necessarily the multiplicative identity.
  $endgroup$
  – Arthur
  Mar 12 at 8:44
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Well, a proof by contradition is not necessary. Suppose $f$ is a unit with inverse $g$. Then $fg=1$.
Using degrees, we obtain
$$0 = rm deg(1) = rm deg(fg) = rm deg(f) + rm deg(g).$$
The last equality holds since $R$ has no zero divisors.
As the degree is a nonnegative function, it follows that both, $f$ and $g$ have degree zero and so are constants (elements of $R$).

share|cite|improve this answer

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

$endgroup$

share|cite|improve this answer

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513

answered Mar 12 at 8:24

WuestenfuxWuestenfux

5,1271513