3
$\begingroup$

I am currently studying commutative algebra and came across the following question.

Let $F$ be a finite field with $q$ elements, let $A=F[x_1,...,x_n]$ and denote by $m$ a maximal ideal in $A$.

  1. How many maximal ideals are in $A$ such that $A/m = F$ ?
  2. How many maximal ideals are in $A$ such that $A/m = L$ , where $|L| = q^k$ ?
  3. How many maximal ideals are in $A$ ?

I know that maximal ideals of $F[x_1,...,x_n]$, where $F$ is an algebraically closed field are of the form $(x-a_1,...,x-a_n)$, but how does a maximal ideal looks like in that kind of a situation?

$\endgroup$
  • $\begingroup$ Use LaTeX please: math.meta.stackexchange.com/questions/5020 $\endgroup$ – Michael Rozenberg Jun 28 at 14:06
  • 1
    $\begingroup$ There are $\frac{1}{k}\sum_{d | k} \mu(d) q^{k/d}$ maximal ideals such that $F_q[x_1]/m = F_{q^k}$. For $n \ge 1$ it is worth looking at the ($\log $ of the ) zeta function of $\Bbb{A}^n_{F_q}$ $\endgroup$ – reuns Jun 28 at 14:17
  • $\begingroup$ Is there another way that you can think of? I'm not familiar with the zeta function @reuns $\endgroup$ – Ariel Jun 28 at 17:15
0
$\begingroup$

Let $f_n(k)$ be the number of morphisms $\Bbb{F}_q[x_1,\ldots,x_n] \to \Bbb{F}_{q^k}$.

They are also morphisms $\Bbb{F}_q[x_1,\ldots,x_n] \to \Bbb{F}_{q^{dk}}$ for every $d$. Thus we can use inclusion exclusion to count the number $g_n(k)$ of them being surjective.

The kernel of such a morphism is a maximal ideal, how many morphisms have the same kernel ?

$\endgroup$
  • $\begingroup$ So, let me see if I understood you right: Let $q=p^n$.Followed by the first isomorphism theorem,we know that there is an isomorphism $\alpha:F_{p^n}[x_1,…,x_n]/ker(\alpha)→F_{p^k}$ where k divides n(because we want the right side to be a field).So,it follows that $ker(α)$ is a maximal ideal.Hence,the number of maximal ideals is equal to the number of surjective homomorphisms(is it just the number of surjective fuctions from A to B in general?)minus the number of surjective homomorphisms which have the same maximal ideal($=p^{k−1}∗p^{k−1}∗p^{k−2}∗...∗1$).@reuns $\endgroup$ – Ariel Jun 29 at 13:22
  • $\begingroup$ @Ariel $f_n(k) = q^{nk}$, $g_n(k) = \sum_{d | k} \mu(d) f_n(k/d)$ and the number of maximal ideals is $\frac{1}{k} g_n(k)$ (because $x\to a, x \to b$ have the same kernel iff $a = (b_1^{q^r},\ldots,b_n^{q^r})$ for some $r$) $\endgroup$ – reuns Jun 29 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.