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Maximal ideals in univariate polynomial rings $R[X]$ have a nice characterization in that they all are of the form $(E)$, for some irreducible $E\in R[X]$. This allows for a systematic way to construct maximal ideals in this setting.

I'm looking to do the same for multivariate polynomial rings. Let $k$ be a field (not algebraically closed -- imagine that it's $\mathbb{F}_p$ for some prime $p$), and let $k[x_1,\ldots,x_n]$ be its polynomial ring in $n$ variables.

More specifically, I'm looking for maximal ideals $I\subseteq k[x_1,\ldots,x_n]$ such that for any polynomial $f\in k[x_1,\ldots,x_n]$ of total degree at most $d$, $f$ is the unique degree $\leq d$ polynomial such that $f = f \mod I$. Otherwise, there exists a degree $\leq d$ polynomial $g$ such that $g = f \mod I$. Intuitively, I would like to have something that behaves like modding by an irreducible $E$ in a univariate polynomial ring: if $E$ is of degree $d+1$, then $f \mod E$ has the behavior I described.

Any other pointers to examples/characterizations of maximal ideals in multivariate polynomial rings would be appreciated!

Thank you!

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I don't understand the question. –  user18119 Dec 14 '12 at 8:44
    
The part "Otherwise...". You are looking for a maximal ideal $I$ satisfying the first part of the condition (before "otherwise") ? –  user18119 Dec 14 '12 at 12:07
    
I edited to hopefully make it more clear. –  Henry Yuen Dec 14 '12 at 17:58
    
By $R[X]$ you meant $\mathbb{R}[X]$? Because the statement isn't true if, for example, $R = \mathbb{Z}[X]$. –  Hurkyl Dec 14 '12 at 18:05
    
I was indeed thinking of R as being something like $\mathbb{F}_p$ or $\mathbb{R}$. However, could you explain why it's not true if $R = \mathbb{Z}[X]$? Thanks! –  Henry Yuen Dec 14 '12 at 18:13

2 Answers 2

up vote 1 down vote accepted

If each $p_i$ is irreducible, then the quotient $k[x_1, ... x_n]/(p_1, ... p_n)$ is canonically isomorphic to the tensor product of fields $\bigotimes k[x_i]/p_i$, so the question is when a tensor product of fields remains a field.

This seems to be a somewhat delicate field-theoretic question in general. Restricting to the case $n = 2$ for simplicity and writing $k_i = k[x_i]/p_i$, note that $k_1 \otimes k_2 \cong k_1[x_2]/p_2$, hence the question is whether $p_2$ remains irreducible when regarded as a polynomial over $k_1$. Then of course one has to repeatedly answer this question for each of the $p_i$.

I think a sufficient condition is that the pairwise intersection of the normal closures of the $k[x_i]/p_i$ in $\bar{k}$ is $k$. (This is wrong; see QiL's comment for the correct condition.) For example one might take $k = \mathbb{Q}$ and $p_i = x_i^2 - q_i$ where $q_i$ is an enumeration of the primes.

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The tensor product is a field is equivalent to the fields $k[x_i]/(p_i)$ are linearly disjoint in an algebraic closure of $k$. –  user18119 Dec 14 '12 at 8:42
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If $k$ is finite, an equivalent condition is that the degrees $\deg p_i$ are pairwise coprime. This is because of the uniqueness of subextensions of given degree in a given algebraic closure. –  user18119 Dec 14 '12 at 12:09
    
Let me try to understand the last comment: if $\deg p_i$ are pairwise coprime, then for a finite field $k$, $\otimes k[x_i]/p_i$ is a field? –  Henry Yuen Dec 14 '12 at 17:59
    
@HenryYuen: yes, this is what I meant. –  user18119 Dec 14 '12 at 18:31

I suspect what you're looking for is the notion of a Groebner basis. The phrase triangular decomposition may be useful.

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