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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 $R[x]$

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2 Answers 2

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.

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Is it ? I am not sure... This does makes sense to me though :O –  Praphulla Koushik Mar 13 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 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.

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i got it.. May be i need some time to digest it.. Thank you.. :) –  Praphulla Koushik Mar 13 at 12:03

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