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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?

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  • $\begingroup$ Use LaTeX please: math.meta.stackexchange.com/questions/5020 $\endgroup$ – Michael Rozenberg Jun 28 at 14:06
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    $\begingroup$ 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
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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 ?

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  • $\begingroup$ 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

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