Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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!

share|cite|improve this question
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
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.

share|cite|improve this answer
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
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.

share|cite|improve this answer

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.