Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Following is an example taken from Dummit Foote - Abstract Algebra after Proposition $9.4.12$

The idea of reducing modulo an ideal to determine irreducibility can be used also in several variables, but some extra care must be exercised. For example, the polynomial $x^2+xy+1$ in $\mathbb{Z}[x,y]$ is irreducible since modulo the ideal $(y)$ it is $x^2+1$ in $\mathbb{Z}[x]$ which is irreducible and of same degree. In this sort of argument it is necessary to be careful about collapsing. For example the polynomial $xy+x+y+1$ (which is $(x+1)(y+1)$) is reducible but appears irreducible modulo both $(x)$ and $(y)$.The reason for this is that non unit polynomials in $\mathbb{Z}[x,y]$ can reduce to units in quotient ring. To take account of this it is necessary to determine which elements in the original ring becomes units in the quotient.

I do not really understand this...

I realize that $xy+x+y+1$ is reducible in $\mathbb{Z}[x,y]$

going modulo $(y)$ gives $x+1$ which is irreducible.

I do not understand what does he mean when he says The reason for this is that non unit polynomials in $\mathbb{Z}[x,y]$ can reduce to units in quotient ring.

$f(x,y)=xy+x+y+1$ after going modulo $(y)$ is $x+1$ which is clearly not a unit...

So, I do not understand what does he actually wants me to see in this.

Please help me to see this.

Thank you.

EDIT : The proposition that i was referring to and which was used in this example is :

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

share|cite|improve this question

I think the other factor is what you are being asked to consider.

Yes, mod $y$ the factor $x+1$ remains irreducible but the factor $y+1$ becomes a unit.

Hence the product $(x+1)(y+1)$ winds up mod $y$ being an irreducible polynomial.

share|cite|improve this answer
Is it ? I am not sure... This does makes sense to me though :O – Praphulla Koushik Mar 13 '14 at 10:37
Note also the consideration of $x^2 + xy + 1$ mod $y$ being $x^2 + 1$ of the same degree. This eliminates the possibility of one factor "collapsing" to a unit, as your example would illustrate. – hardmath Mar 13 '14 at 12:45

You need to remember the proof of the proposition you quote. Basically suppose $p(x)$ is reducible in $R[x]$, then $p(x) = q(x) r(x)$ where neither $q$ nor $r$ is a unit (of $R$). When you reduce $\bar{p}(x) = p(x) \pmod{I}$, you get $\bar p = \bar q \bar r$. And the key fact is that if a polynomial in $R[x]$ is not a unit, then its reduction mod I is not a unit. So neither $\bar q$ nor $\bar r$ is a unit, and $\bar p$ is reducible.

This breaks down in $R[x,y]$, because $y+1$ is not a unit, but its reduction mod $(y)$ is. So when you try to apply the same proof to $p = xy+x+y+1 = (x+1)(y+1) = q r$, the second factor $r$ is a unit mod $(y)$. So you're left with only one factor $\bar p = \bar q$. It's not the reduction of the whole polynomial that matters: it's the reduction of each factor.

share|cite|improve this answer
i got it.. May be i need some time to digest it.. Thank you.. :) – Praphulla Koushik Mar 13 '14 at 12:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.