Let $K$ be a field and $R$ a commutative ring that's a finite dimensional $K$-vector space.
Do we necessarily have (as rings) $$ R \cong \Pi_{i=1}^n F_i $$ for some finite field extensions $F_i \supseteq K$, with the usual product ring multiplication? Is there a proof of this or a counter example?
I know that $R$ is a field if and only if $R$ is a domain, so to have this not be the case a ring would need "sufficient" zero-divisors.
I was thinking about other possible multiplicative structures, and thought of those direct sums of products after a bit ($K^n$ or direct sums of finite field extensions). Is it possible to have other structures with less "obvious" multiplication?
I was trying to come up with a few, but the role of the identity seems to be troubling me:
I tried constructing one on $\mathbb{R}^2$ where $$ (a,b) \cdot (c,d) = (a^2c^2,b^2d^2) $$ or something with each component multiplying non-trivially on the others, but then if the identity is $(e,f)$, then it doesn't act trivially: $$ (a,b) = (a,b) \cdot (e,f) = (a^2e^2,b^2f^2) \implies e^2=a^{-1} $$
The additive group will be isomorphic to $K^n$ by definition, so I'm thinking $(1,\dots,1)$ might always somehow be an identity element, at least up to an automorphism of $K^n$ for instance maybe $\mathbb{F}_p$ could somehow multiply through an additive homomorphism: $$ \phi : \ \mathbb{F}_p \to \mathbb{F}_p \ : \ 1 \mapsto a $$ then let $$ b * c := \phi\bigg(\phi^{-1}(b)\phi^{-1}(c)\bigg) $$ but then this is just an automorphism so it's still a field, so $\mathbb{F}_p^2$ with one component the usual multiplication, and one twisted (is this the right word?) by $\phi$, would be isomorphic to $\mathbb{F}_p^2$ with the usual ring structure.
Maybe some sort of intertwining, on $\mathbb{R}^2$ for instance $$ (a,b) \cdot (c,d) = (ac,bcd) $$ (where here if $(a,b)$ is the identity it wouldn't work).
I'm not sure where to start on a proof of non-existence either... Any help greatly appreciated!
In general, no: The ring $R$ can have nilpotent elements, but any direct sum $\bigoplus F_i$ of fields $F_i$ cannot. The simplest example, mentioned by paul garrett in the comments, is:
Example For any field $K$ and $n > 1$, take $$R := K[x] / \langle x^n \rangle$$ and denote the image of $x$ under the projection $\Bbb F[x] \to \Bbb F[x] / \langle x^n \rangle$ by $\alpha$. Then $\alpha, \ldots, \alpha^{n - 1}$ are nilpotent.
If $K$ is finite, then $R$ can also have counts of zero divisors not achievable by any sum of fields. For example, if $K = \Bbb F_2$, $n = 2$, the ring $$R = \Bbb F_2[x] / \langle x^2 \rangle$$ has $1$ zero divisor ($x$), but the only sums of fields extending $\Bbb F_2$ of dimension $2$ over $\Bbb F_2$ are