Isomorphic to $\mathbb{Q}$ ? Part II.

By the Eisenstein criterion, $f(x) = \displaystyle \sum_{k=0}^{2026} x^k$ is irreducible. (This is quite a nice problem, actually. The trick is, it suffices to show that $f(x+1)$ is irreducible (and you can look at the solution to the previous problem to see why this is so)). So $\mathbb{Q}[x]/(f(x))$ is a field. Now it suffices to show that $\mathbb{Q}[x]/(f(x))$ contains no smaller subfields. Clearly, $\mathbb{Q}[x]/(f(x))$ is countably infinite (Use the following lemmas: Let $R$ be a commutative ring, and $f(x)$ be a monic polynomial with degree $n \geqslant 1$ . Then every element of $R/(f(x))$ is of the form $\overline{p(x)}, p(x) \in R[x]$ and $\deg p < n$ , and if $p, q \in R[x]$ are distinct polynomials of degree $< n$ , then $\overline{p(x)} \neq \overline{q(x)}$ ; the first lemma is an induction proof and the second is relatively easy ). So $\mathbb{Q}[x]/(f(x))$ has no finite subfields, and the smallest subfield of an infinite field is isomorphic to $\mathbb{Q}$ , which is countably infinite. This completes the proof. $\square$