Why isn't this criterion for determining irreducibilty working?

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I have learned this criterion for irreducibility of polynomials:

Let $R$ be an integral domain, let $I$ be a proper ideal of $R$, and let $p(x)$ be a non-constant monic polynomial in $R[x]$. If the image of $p(x)$ is irreducible in $(R/ I)[x]$ under the natural homomorphism, then $p(x)$ is irreducible in $R[x]$.

Now the polynomial $xy+x+y+1 = (x+1)(y+1)$ is reducible in $mathbbZ[x, y] = (mathbbZ[y])[x]$. Take $R = mathbbZ[y]$ and $I = (y)$. Then the image of $xy+x+y+1$ in $(mathbbZ[y]/(y))[x]$ is $x+1$, which is irreducible because $mathbbZ[y]/(y) cong mathbbZ$ and $x+1$ is irreducible in $mathbbZ[x]$. Why does this criterion seem to not work in this situation?

Edit: Proof of the criterion

We prove the contrapositive. Suppose $p(x)$ is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.

abstract-algebra ring-theory polynomial-rings

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edited Jul 25 at 20:06

asked Jul 25 at 19:46

Ovi

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find that criterion?
  – Kenny Lau
  Jul 25 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
  – Ovi
  Jul 25 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Add that to the question.
  – Kenny Lau
  Jul 25 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau Done $$
  – Ovi
  Jul 25 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
  – Kenny Lau
  Jul 25 at 20:06
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

I have learned this criterion for irreducibility of polynomials:

Let $R$ be an integral domain, let $I$ be a proper ideal of $R$, and let $p(x)$ be a non-constant monic polynomial in $R[x]$. If the image of $p(x)$ is irreducible in $(R/ I)[x]$ under the natural homomorphism, then $p(x)$ is irreducible in $R[x]$.

Now the polynomial $xy+x+y+1 = (x+1)(y+1)$ is reducible in $mathbbZ[x, y] = (mathbbZ[y])[x]$. Take $R = mathbbZ[y]$ and $I = (y)$. Then the image of $xy+x+y+1$ in $(mathbbZ[y]/(y))[x]$ is $x+1$, which is irreducible because $mathbbZ[y]/(y) cong mathbbZ$ and $x+1$ is irreducible in $mathbbZ[x]$. Why does this criterion seem to not work in this situation?

Edit: Proof of the criterion

We prove the contrapositive. Suppose $p(x)$ is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.

abstract-algebra ring-theory polynomial-rings

share|cite|improve this question

edited Jul 25 at 20:06

asked Jul 25 at 19:46

Ovi

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find that criterion?
  – Kenny Lau
  Jul 25 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
  – Ovi
  Jul 25 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Add that to the question.
  – Kenny Lau
  Jul 25 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau Done $$
  – Ovi
  Jul 25 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
  – Kenny Lau
  Jul 25 at 20:06
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

up vote
4
down vote

favorite

I have learned this criterion for irreducibility of polynomials:

Let $R$ be an integral domain, let $I$ be a proper ideal of $R$, and let $p(x)$ be a non-constant monic polynomial in $R[x]$. If the image of $p(x)$ is irreducible in $(R/ I)[x]$ under the natural homomorphism, then $p(x)$ is irreducible in $R[x]$.

Now the polynomial $xy+x+y+1 = (x+1)(y+1)$ is reducible in $mathbbZ[x, y] = (mathbbZ[y])[x]$. Take $R = mathbbZ[y]$ and $I = (y)$. Then the image of $xy+x+y+1$ in $(mathbbZ[y]/(y))[x]$ is $x+1$, which is irreducible because $mathbbZ[y]/(y) cong mathbbZ$ and $x+1$ is irreducible in $mathbbZ[x]$. Why does this criterion seem to not work in this situation?

Edit: Proof of the criterion

We prove the contrapositive. Suppose $p(x)$ is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.

abstract-algebra ring-theory polynomial-rings

share|cite|improve this question

edited Jul 25 at 20:06

asked Jul 25 at 19:46

Ovi

11.3k935105

I have learned this criterion for irreducibility of polynomials:

Let $R$ be an integral domain, let $I$ be a proper ideal of $R$, and let $p(x)$ be a non-constant monic polynomial in $R[x]$. If the image of $p(x)$ is irreducible in $(R/ I)[x]$ under the natural homomorphism, then $p(x)$ is irreducible in $R[x]$.

Now the polynomial $xy+x+y+1 = (x+1)(y+1)$ is reducible in $mathbbZ[x, y] = (mathbbZ[y])[x]$. Take $R = mathbbZ[y]$ and $I = (y)$. Then the image of $xy+x+y+1$ in $(mathbbZ[y]/(y))[x]$ is $x+1$, which is irreducible because $mathbbZ[y]/(y) cong mathbbZ$ and $x+1$ is irreducible in $mathbbZ[x]$. Why does this criterion seem to not work in this situation?

Edit: Proof of the criterion

We prove the contrapositive. Suppose $p(x)$ is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.

abstract-algebra ring-theory polynomial-rings

share|cite|improve this question

edited Jul 25 at 20:06

asked Jul 25 at 19:46

Ovi

11.3k935105

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 20:06

edited Jul 25 at 20:06

edited Jul 25 at 20:06

asked Jul 25 at 19:46

Ovi

11.3k935105

asked Jul 25 at 19:46

Ovi

11.3k935105

asked Jul 25 at 19:46

Ovi

11.3k935105

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find that criterion?
  – Kenny Lau
  Jul 25 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
  – Ovi
  Jul 25 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Add that to the question.
  – Kenny Lau
  Jul 25 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau Done $$
  – Ovi
  Jul 25 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
  – Kenny Lau
  Jul 25 at 20:06
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find that criterion?
  – Kenny Lau
  Jul 25 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
  – Ovi
  Jul 25 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Add that to the question.
  – Kenny Lau
  Jul 25 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau Done $$
  – Ovi
  Jul 25 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
  – Kenny Lau
  Jul 25 at 20:06
  

  
  
  
  

Where did you find that criterion?
– Kenny Lau
Jul 25 at 20:00

Where did you find that criterion?
– Kenny Lau
Jul 25 at 20:00

@KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
– Ovi
Jul 25 at 20:05

@KennyLau I forgot to add that $p(x)$ is non-constant and monic. My friend told me this criterion, and the contrapositive has an easy proof: Suppose is reducible, $p(x) = a(x)b(x)$. Since $p(x)$ is monic, $a(x)$ and $b(x)$ are non-constant and monic. Thus when you reduce the coefficients $pmod I$ you get a factorization in $(R/I)[x]$.
– Ovi
Jul 25 at 20:05

Add that to the question.
– Kenny Lau
Jul 25 at 20:05

Add that to the question.
– Kenny Lau
Jul 25 at 20:05

@KennyLau Done $$
– Ovi
Jul 25 at 20:06

@KennyLau Done $$
– Ovi
Jul 25 at 20:06

@GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
– Kenny Lau
Jul 25 at 20:06

@GalPorat $x+1$ and $y+1$ are non-units, so that is a perfectly valid proof that the polynomial is reducible.
– Kenny Lau
Jul 25 at 20:06

 | 
show 2 more comments

1 Answer
1

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votes

up vote
7
down vote



accepted

$xy + x + y + 1 in (Bbb Z[y])[x]$ is not monic, for its leading coefficient is $y+1$.

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohhhhh thanks! $$
  – Ovi
  Jul 25 at 20:07
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

$xy + x + y + 1 in (Bbb Z[y])[x]$ is not monic, for its leading coefficient is $y+1$.

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohhhhh thanks! $$
  – Ovi
  Jul 25 at 20:07
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

$xy + x + y + 1 in (Bbb Z[y])[x]$ is not monic, for its leading coefficient is $y+1$.

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohhhhh thanks! $$
  – Ovi
  Jul 25 at 20:07
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

$xy + x + y + 1 in (Bbb Z[y])[x]$ is not monic, for its leading coefficient is $y+1$.

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

$xy + x + y + 1 in (Bbb Z[y])[x]$ is not monic, for its leading coefficient is $y+1$.

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

answered Jul 25 at 20:05

Kenny Lau

18.5k2157

18.5k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohhhhh thanks! $$
  – Ovi
  Jul 25 at 20:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohhhhh thanks! $$
  – Ovi
  Jul 25 at 20:07
  

  
  
  
  

Ohhhhh thanks! $$
– Ovi
Jul 25 at 20:07

Ohhhhh thanks! $$
– Ovi
Jul 25 at 20:07

add a comment | 

 

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