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Take a finite field $\mathbb{F}_p$. To construct $\mathbb{F}_{p^n}$ one takes the field of degree $d < n$ polynomials over $\mathbb{F}_q$ with operations done mod an irreducible degree $n$ polynomial $P(X)$.

What happens when we take $P(X)$ to be reducible?

In particular, let $P(X) = f_1^{e_1} \dots f_k^{e_k}$ with $f_i$ distinct and irreducible. If all $e_1 = e_2 \dots = e_k = 1$ I suppose using CRT you would get the product of finite fields. What happens when $e_i > 1$? You get nilpotent elements (e.g. $f_i$) so it certainly can't be a product of finite fields. Does it depend on the choice of $f_i$?

MT_
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  • If you think of $\Bbb F_p[x]$ geometrically as $\Bbb A^1_{\Bbb F_p}$, you can take the point of view that this is something like $\Bbb F_p$, along with "extra points" for field extensions of $\Bbb F_p$. In reality, $\Bbb A^1_{\Bbb F_p}$ can be seen as $\overline{\Bbb F}_p\cup{\eta}$, modulo the equivalence relation $x\sim y$ ($x,y\in\overline{\Bbb F}_p$) if there exists $\sigma\in\operatorname{Gal}(\overline{\Bbb F}_p/\Bbb F_p)$ such that $\sigma(x) = y$ (i.e., $x$ and $y$ have the same minimal polynomial over $\Bbb F_p$), and $\eta$ is a generic point that is somehow "everywhere." ctd – Stahl Nov 11 '16 at 21:03
  • ctd The quotient ring $\Bbb F_p[x]/(f_1^{e_i}\dots f_n^{e_n})$ can be seen as a subset $Z$ of $\Bbb A^1_{\Bbb F_p}$ by $[x]\in Z$ if $f_i(x) = 0$ for some $i$. This hasn't captured the nilpotentcy, but you might view the points of the subset being sort of fuzzy (you get more fuzz if you have a larger $e_i$). In reality, the nilpotency is realized by nilpotent elements in the structure sheaf of $Z$, but we remember this in picture form as fuzz. (This might not be helpful, as it simply recasts $\Bbb F_p[x]/(f_i^{e_i})$ in the role of the structure sheaf of $Z$ and doesn't explicitly describe it) – Stahl Nov 11 '16 at 21:03
  • See http://math.stackexchange.com/questions/305824/structure-of-finite-commutative-rings. – lhf Nov 11 '16 at 21:23

2 Answers2

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The Chinese remainder theorem says the canonical map: $$\mathbf F_p[x]/(f_1^{e_i}\dots f_n^{e_n})\simeq \mathbf F_p[x]/(f_1^{e_1})\times\dots\times\mathbf F_p[x]/(f_n^{e_n})$$ is an isomorphism.

Each factor $\;\mathbf F_p[x]/(f_i^{e_i})\;$ is an artinian local $k$-algebra, with maximal ideal $\;(f_i)/(f_i^{e_i})$.

Bernard
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Consider e.g. $R = \mathbb F[X]/(f(X)g(X))$ for irreducible polynomials $f(X)$, $g(X)$. This is a ring, but $f(X)$ and $g(X)$ are zero-divisors, so in particular it is not a field.

Robert Israel
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