Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi. Given $k$ an algebraically closed field, I know that that a maximal ideal $\mathfrak{m}$ of $A = k[X_1,\cdots,X_n]$ is just a $\langle X_1-a_1,\cdots,X_n-a_n \rangle $ (Nullstellensatz). Knowing that, it seems intuitive that $\mathfrak{m}$ can not be generated by less than $n$ elements. Is that true ? In that case, how can I show that ?

(Actually, I need that, just after using Nakayama's lemma, to show that $dim_{A/\mathfrak{m}=k} \mathfrak{m}/\mathfrak{m}^2$ is greater than $n$.)

(I've seen things that might be relevant like "Krull height theorem" but I think such things take place in a more general context and I have some difficulties, first to understand them, and second to adapt them...)

Thank you.

share|improve this question

4 Answers 4

up vote 5 down vote accepted

Since you mentioned Krull's height theorem (= the generalized principal ideal theorem) and having difficulty applying it, I thought you or someone else might appreciate seeing how this works: it is quite straightforward.

The generalized principal ideal theorem is as follows: let $R$ be a Noetherian ring and $I$ a proper ideal of $R$ which can be generated by $n$ elements. Let $\mathfrak{p}$ be a prime ideal which is minimal among all primes containing $I$. Then $\mathfrak{p}$ has height at most $n$, that is, there do not exist prime ideals $\mathfrak{p}_0,\ldots,\mathfrak{p}_n$ such that

$\mathfrak{p}_0 \subsetneq \mathfrak{p_1} \subsetneq \ldots \subsetneq \mathfrak{p_n} \subsetneq \mathfrak{p}$.

[For a deduction of this from Krull's Principal Ideal Theorem, see e.g. Theorem 96 on p. 70 of http://math.uga.edu/~pete/integral.pdf.]

Let us apply this with $R = k[x_1,\ldots,x_n]$ and the ideal $I = \langle x_1 - a_1,\ldots,x_n - a_n \rangle$. $I$ is itself a maximal -- hence prime -- ideal, since $R/I \cong k$. Thus the generalized principal ideal theorem simply says that $I$ cannot be generated by fewer elements than its height. But its height is certainly at least $n$. No geometry is needed here: just define $\mathfrak{p}_0 = 0$ and for $1 \leq i \leq n$, $\mathfrak{p}_i = \langle x_1 - a_1,\ldots,x_i - a_i \rangle$.

Finally, a comment: I did not use that $k$ was algebraically closed per se but only worked with maximal ideals of this particular form. On the other hand, it is still true over an arbitrary field $k$ that every maximal ideal of $k[x_1,\ldots,x_n]$ has height $n$ and can be generated by $n$ elements (and no fewer, by Krull's theorem): see Corollary 130 on p. 83 of the document linked to above.

share|improve this answer
    
Thanks a lot, that exactly answers my question ! –  Laurent May 25 '10 at 18:22

The classes of the $X_i - a_i$ are easily seen to be a basis for $\mathfrak{m}/\mathfrak{m}^2$.

share|improve this answer

The dimension is exactly n: you are counting linear polynomials modulo terms of higher order.

The geometrical interpretation is that intersecting n - 1 hypersurfaces with a common point must give an algebraic set with a component of dimension at least one. If you insist on a commutative algebra approach, you are going to have to study those theorems ...

share|improve this answer

Trying to write down something along the lines you were looking for, I would argue that over an algebraically closed field the set of common zeros of $k < n$ polynomials has codimension at most $k$ and so cannot be a single point $(a_1,\ldots,a_n)$. You can use Krull theorem indeed (e.g. check out Ravi Vakil's notes: http://math.stanford.edu/~vakil/725/class14.pdf), or argue more along the 19th century and prove by some sort of elimination that the set of common zeros depend on at least $n-k$ parameters...

share|improve this answer

Your Answer

 
discard

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.