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Is it true that no polynomial (with the exception of constant polynomial) have an inverse in $mathbb Z/pmathbb Z$ (where p is prime)?

0

$begingroup$

I was solving the following exercise:

Find the inverse of $p(x) = 1 + x$ in $R[x]$ over $mathbb Z/5mathbb Z$ or show that it does not exist.

and finding that it does not exist because if there is a polynomial $q(x)$ such that $p(x)q(x)=1$. Which will contradict the proposition:

Let $R$ be a commutative ring. For every pair of non-zero polynomials $p$ and $q$ in $R[x]$, if the leading coefficient of $p$ or $q$ is not a zero divisor in $R$ then $deg(pq)=deg(p)+deg(q)$

my question is it safe to assume that there does not exist an inverse of polynomial of $deg>0$ over $mathbb Z/pmathbb Z$ where $p$ is a prime number?

polynomials ring-theory finite-fields

share|cite|improve this question

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused. You seem to have proved a statement while, at the same time, you are asking whether you can safely assume that the statment is true.
  $endgroup$
  – José Carlos Santos
  Apr 2 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does "inverse" mean here? Do you mean "reciprocal"? Then, sure. Do you mean "inverse function"? Well, then $q(x)=x-1$ works.
  $endgroup$
  – lulu
  Apr 2 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lulu I think it's clear from "$p(x)q(x)=1$" that OP means multiplicative inverse.
  $endgroup$
  – cansomeonehelpmeout
  Apr 2 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @cansomeonehelpmeout Well, I think it's clear that the OP approached the problem from that point of view. But was that what was intended? Seems an odd phrasing. Why multiplicative inverse? Additive inverse is simpler.
  $endgroup$
  – lulu
  Apr 2 at 15:13
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I was solving the following exercise:

Find the inverse of $p(x) = 1 + x$ in $R[x]$ over $mathbb Z/5mathbb Z$ or show that it does not exist.

and finding that it does not exist because if there is a polynomial $q(x)$ such that $p(x)q(x)=1$. Which will contradict the proposition:

Let $R$ be a commutative ring. For every pair of non-zero polynomials $p$ and $q$ in $R[x]$, if the leading coefficient of $p$ or $q$ is not a zero divisor in $R$ then $deg(pq)=deg(p)+deg(q)$

my question is it safe to assume that there does not exist an inverse of polynomial of $deg>0$ over $mathbb Z/pmathbb Z$ where $p$ is a prime number?

polynomials ring-theory finite-fields

share|cite|improve this question

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused. You seem to have proved a statement while, at the same time, you are asking whether you can safely assume that the statment is true.
  $endgroup$
  – José Carlos Santos
  Apr 2 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does "inverse" mean here? Do you mean "reciprocal"? Then, sure. Do you mean "inverse function"? Well, then $q(x)=x-1$ works.
  $endgroup$
  – lulu
  Apr 2 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lulu I think it's clear from "$p(x)q(x)=1$" that OP means multiplicative inverse.
  $endgroup$
  – cansomeonehelpmeout
  Apr 2 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @cansomeonehelpmeout Well, I think it's clear that the OP approached the problem from that point of view. But was that what was intended? Seems an odd phrasing. Why multiplicative inverse? Additive inverse is simpler.
  $endgroup$
  – lulu
  Apr 2 at 15:13
  
  

  
  
  
  

add a comment | 

$begingroup$

I was solving the following exercise:

Find the inverse of $p(x) = 1 + x$ in $R[x]$ over $mathbb Z/5mathbb Z$ or show that it does not exist.

and finding that it does not exist because if there is a polynomial $q(x)$ such that $p(x)q(x)=1$. Which will contradict the proposition:

Let $R$ be a commutative ring. For every pair of non-zero polynomials $p$ and $q$ in $R[x]$, if the leading coefficient of $p$ or $q$ is not a zero divisor in $R$ then $deg(pq)=deg(p)+deg(q)$

my question is it safe to assume that there does not exist an inverse of polynomial of $deg>0$ over $mathbb Z/pmathbb Z$ where $p$ is a prime number?

polynomials ring-theory finite-fields

share|cite|improve this question

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

$endgroup$

share|cite|improve this question

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

share|cite|improve this question

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

edited Apr 2 at 15:02

cansomeonehelpmeout

7,3323935

asked Apr 2 at 14:51

KbiirKbiir

696

asked Apr 2 at 14:51

KbiirKbiir

696

2 Answers
2

active

oldest

votes

2

$begingroup$

Yes it is true, this is because $deg(PQ)=deg(P)+deg(Q)$ for nonzero polynomials, so both summands must be $0$.

share|cite|improve this answer

edited Apr 2 at 15:02

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

$endgroup$

add a comment | 

1

$begingroup$

Well, for any polynomials $f,gin R[x]$, where $R$ is an integral domain (i.e., has no zero divisors), the degree formula
$$deg(fcdot g) = deg(f)+deg(g)$$
holds, since the highest terms do not cancel.

Thus if $f$ is a unit in $R[x]$, then $fcdot g =1$ for some $gin R[x]$.
Then $0 =deg (1) = deg(fcdot g) =deg(f) + deg(g)$.
Since the degrees are nonnegative it follows that $deg(f)=0$ and so $f$ is invertible in $R$. Done.

share|cite|improve this answer

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

$endgroup$

add a comment | 

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2

$begingroup$

Yes it is true, this is because $deg(PQ)=deg(P)+deg(Q)$ for nonzero polynomials, so both summands must be $0$.

share|cite|improve this answer

edited Apr 2 at 15:02

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

$endgroup$

add a comment | 

2

$begingroup$

Yes it is true, this is because $deg(PQ)=deg(P)+deg(Q)$ for nonzero polynomials, so both summands must be $0$.

share|cite|improve this answer

edited Apr 2 at 15:02

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

$endgroup$

add a comment | 

$begingroup$

Yes it is true, this is because $deg(PQ)=deg(P)+deg(Q)$ for nonzero polynomials, so both summands must be $0$.

share|cite|improve this answer

edited Apr 2 at 15:02

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

$endgroup$

share|cite|improve this answer

edited Apr 2 at 15:02

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

answered Apr 2 at 14:55

Jorge Fernández HidalgoJorge Fernández Hidalgo

77.2k1394195

1

$begingroup$

Well, for any polynomials $f,gin R[x]$, where $R$ is an integral domain (i.e., has no zero divisors), the degree formula
$$deg(fcdot g) = deg(f)+deg(g)$$
holds, since the highest terms do not cancel.

Thus if $f$ is a unit in $R[x]$, then $fcdot g =1$ for some $gin R[x]$.
Then $0 =deg (1) = deg(fcdot g) =deg(f) + deg(g)$.
Since the degrees are nonnegative it follows that $deg(f)=0$ and so $f$ is invertible in $R$. Done.

share|cite|improve this answer

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

$endgroup$

add a comment | 

1

$begingroup$

Well, for any polynomials $f,gin R[x]$, where $R$ is an integral domain (i.e., has no zero divisors), the degree formula
$$deg(fcdot g) = deg(f)+deg(g)$$
holds, since the highest terms do not cancel.

Thus if $f$ is a unit in $R[x]$, then $fcdot g =1$ for some $gin R[x]$.
Then $0 =deg (1) = deg(fcdot g) =deg(f) + deg(g)$.
Since the degrees are nonnegative it follows that $deg(f)=0$ and so $f$ is invertible in $R$. Done.

share|cite|improve this answer

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

$endgroup$

add a comment | 

$begingroup$

Well, for any polynomials $f,gin R[x]$, where $R$ is an integral domain (i.e., has no zero divisors), the degree formula
$$deg(fcdot g) = deg(f)+deg(g)$$
holds, since the highest terms do not cancel.

Thus if $f$ is a unit in $R[x]$, then $fcdot g =1$ for some $gin R[x]$.
Then $0 =deg (1) = deg(fcdot g) =deg(f) + deg(g)$.
Since the degrees are nonnegative it follows that $deg(f)=0$ and so $f$ is invertible in $R$. Done.

share|cite|improve this answer

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

$endgroup$

share|cite|improve this answer

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513

answered Apr 2 at 15:03

WuestenfuxWuestenfux

5,5911513