Let $F$ be a field and $R=\operatorname{End}_{F}(F[x])$. Prove or disprove that $R^2\cong R$ as rings.

Question

Let $F$ be a field and $R=\operatorname{End}_{F}(F[x])=\operatorname{Hom}_{F}(F[x],F[x])$. Prove or disprove that $R^2\cong R$ as rings.

This question is from my homework. The original question only asked me to prove it, but then my professor modified it by changing "as rings" to "as $R$-module", which made the problem much simpler. However, this made me curious whether the original statement was correct.
To find an $R$-module isomorphism, I construct $\Phi:R\to R^2$ by $\Phi(f)=(\pi_{1}\circ f,\pi_{2}\circ f)$, where $\pi_{k}\in \operatorname{End}_{F}(F[x]) $ by $\pi_{1}(x^{2n})=x^n$, $\pi_{1}(x^{2n+1})=0$, and $\pi_{2}$ is defined in the similar way. But under this construction, $\Phi$ seems not to be a ring isomorphism.

Edit: My question is about proving the ring isomorphism, not the R-module isomorphism!

Answer 1

The thing about $F[x]$ here is that it's just a fancy way of writing a countable dimensional $F$ vector space $V$. It's known that for any vector space $V_F$, $End(V_F)$ is a prime ring, which is to say two nonzero ideals cannot multiply to zero.

This is apparent from the fact that $V$ is a faithful $End(V_F)$ module, but you can also easy prove it from scratch that if $I$ is a nonzero ideal, then it necessarily contains all rank $1$ projections. These are idempotents, and it is easy to show you can't annihilate them all at once with something nonzero.

If $R\cong R\times R$ as rings, then the inverse images of $R\times\{0\}$ and $\{0\}\times R$ are two nonzero ideals that multiply to zero.

Answer 2

Note that $R^2$ has central orthogonal idempotents $(0,1)$ and $(1,0)$.

Thus, if $R \cong R^2$, then $R$ has central orthogonal idempotents $e_1, e_2 \in R$ such that $e_1 + e_2 = 1$.

In particular, $e_1, e_2 : F[x] \to F[x]$ would be central! However, the only central $F$-linear endomorphisms of $F[x]$ are "multiplication by $\lambda$" for some $\lambda \in F$.

Then $e_1, e_2$ being idempotents would imply that $e_1 = e_2 = 1$, but this contradicts orthogonality!

Thus, $R \ncong R^2$ as rings.