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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

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Perhaps we should begin by discussing the proof of the proposition in detail. If we emphasize a particularly important step, we will then see why when we consider $\mathbb{Z}[x,y]$, we are lead to the remark that you're asking about.

The proof of the proposition (paraphrasing):

Assume $p(x)$ is reducible in $R[x]$ but irreducible in $(R/I)[x]$. Then $p(x)=f(x)g(x)$ with $f(x), g(x)$ monic, non-constant polynomials. Then reducing $f(x)$ and $g(x)$ mod $I$ gives a factorization in $(R/I)[x]$. Contradiction.

Notice that it was critical that both $f(x)$ and $g(x)$ were monic (or at least that their leading coefficients are units, which can be shown to be equivalent to choosing them to be monic). Why? Because otherwise $f(x)$ could reduce to a unit in $(R/I)[x]$ (this could happen easily, think of $3x+1 \in \mathbb{Z}[x]$ and then consider $\mathbb{Z}_3[x]$). Then we wouldn't have found found that $p(x)$ was reducible in $(R/I)[x]$ and there would be no contradiction!

For completeness: the key point in this proposition is that we are assuming that $p(x)$ is monic and $I$ is a proper ideal. Then the leading coefficients of $f(x)$ and $g(x)$ are units (we argue that we can always multiply by a constant to arrange the leading coefficients to be $1$) and since $I$ is a proper ideal, these coefficients are not in $I$. Then the leading coefficients of $f(x)$ mod $I$ and $g(x)$ mod $I$ are not zero and so we have a nontrivial factorization in $(R/I)[x]$.

So really we need to emphasize why $f(x)$ and $g(x)$ couldn't reduce to a unit. Then it becomes more clear why we need to also worry about factors reducing to units when we consider polynomial rings of several variables.


Some further remarks:

Let's say we try to extend this proposition to $R[x,y]$. But we immediately have a problem: we know we need some kind of "leading coefficient" condition if we hope to repeat the proof as before (or else we need some other condition and strategy to show that reducible $q(x,y)$ cannot become irreducible in a quotient by a proper ideal!). But what we mean by "leading coefficient" is unclear for something like

\begin{align} q(x,y) = x^3y-6xy^3+x^2y^2+x^2y+3xy^2+x-2y+1 \end{align}

Maybe we mean coefficients of terms of highest degree ($x^3y-6xy^3+x^2y^2$ in this case). Then is it enough to ask that at least one of them have coefficient $1$? All of them? But then notice that

\begin{align} q(x,y) = (x^2y+3xy^2+1)(x-2y+1) \end{align}

and so one of the maximum degree terms $x^2y^2$ doesn't arise as simply the product of two lower degree terms - it is the sum of two products. So $x^2y^2$ having coefficient $1$ didn't mean that it couldn't come from terms that are $0$ mod some ideal. The situation can get worse for higher degrees and higher numbers of variables: terms with coefficient 1 could come from a large number of sums of terms. This means we have to think carefully before trying to apply our previous proof.

One such strategy is to recognize that $R[x,y] \cong (R[x])[y]$. In other words the multivariable case can be thought of as the single variable case with coefficients coming from the ring $R[x]$ and so then the condition is simply that the term with the highest power of $x$ (or $y$) should be monic (more generally: check that it's enough to require that the coefficient of the highest power of say $y$ is not an element of the ideal $I[x] \subset R[x]$ that you want to reduce by). If this is the case, then we can simply apply the original proposition of the single variable case. For example

\begin{align} p(x,y) = y^2 + (7x^2+14x)y+17 \end{align}

has this property and so we have $p(x,y) = y^2 + 3$ mod $7$, which is irreducible and so $p(x,y)$ was irreducible.

Notice that Dummit and Foote's example of $xy+x+y+1$ does not meet this criteria (neither the coefficient of the highest power of $x$ nor $y$ is $1$), so while this wouldn't have proved it was reducible, it would have prevented us from making the mistake they warn against. This criteria also would have been applicable to their other example $x^2 + xy+1$ (and we would have found that it is irreducible).

Since they have not yet introduced the idea of reducing mod some $I[x,y]$, hopefully combining the two parts of my answer fully resolves what we're trying to do, why we can do it, and where we can't, at least with respect to the context of what they're trying to point out at this point in the book.

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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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  • $\begingroup$ Is it ? I am not sure... This does makes sense to me though :O $\endgroup$
    – user87543
    Mar 13, 2014 at 10:37
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    $\begingroup$ 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. $\endgroup$
    – hardmath
    Mar 13, 2014 at 12:45

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