$C(X,K)$ isomorphic to polynomial algebra $K[t]$

Question

For $X$ a topological space and $K$ a topological field of characteristic $0$, let $C(X,K)$ be the set of all continuous functions $f:X\to K$. The set $C(X,K)$ turns to a ring with point-wise addition and multiplication: \begin{align} (f+g)(x)&=f(x)+g(x)\\ (fg)(x) & =f(x)g(x). \end{align} I'm looking for a topological space $X$ such that $C(X,K)\cong K[t]$. Or I'm looking for a proof that shows such space does not exist.

Thanks.

Answer 1

This is not a full answer, but just leads. First of all, as discussed in the comments, I'll consider that we're asking for a $K$-algebra isomorphism, not just an isomorphism as abstract rings.

Note that if $X$ is such a space, the $K$-algebra structure map is $\lambda \mapsto $ the constant map with value $\lambda$.

Now, if $f:X\to K$ is continuous and nonconstant, then under the isomorphism $C(X,K)\to K[t]$ it is sent to a nonconstant polynomial, hence to a noninvertible element of the algebra.

But now we see that if $f:X\to K$ doesn't vanish, i.e. factors through $K^\times$, then it's invertible in $C(X,K)$, thus in $K[t]$, thus it must be a constant map:

there are only constant maps $X\to K^\times$.

Now let $f$ be the antecedent of $t$. In particular, for all non constant $P\in K[t]$, $P(f)$ is sent to $P(t)$, thus is noninvertible; and therefore it has a zero, say $x$. Then $f(x)$ is a root of $P$: every nonconstant polynomial has a root, which means that $K$ is algebraically closed.

If such an $X$ exists, $K$ is algebraically closed.

Notice that with the same reasoning, we get that for all $\lambda$, $f-\lambda$ is non invertible, thus $f$ is surjective.

There is a surjective continuous map $X\to K$ which generates $C(X,K)$.

Now if we have a nonconstant continuous map $K\to K^\times$, putting the first and last yellow results together, we get a contradiction. This is the case for instance with $K=\mathbb{C}$ with the usual topology, where $\exp$ is continuous and nonconstant.

There is no such space for $\mathbb{C}$. More generally, for every topological algebraically closed field with a continuous nonconstant map $K\to K^\times$, there is no such $X$.

I don't know whether all topological algebraically closed fields have such a map. What about $\overline{\mathbb{Q}}$ for instance ?

In any case, it feels like the constraint "there is no nonconstant continuous $g:X\to K^\times$" is a pretty big one; but I don't know how to make that precise yet. The isomorphism with $K[t]$ seems to suggest that $X$ doesn't have many opens, but this constraint suggests that it has a lot...