Suppose $F ⊂ K$ are fields. Let $f(x) ∈ F[x] ⊂ K[x]$. Suppose that $f(x)$ is irreducible in $K[x]$. Prove that $f(x)$ is also irreducible in $F[x]$. Suppose $F ⊂ K$ are both fields. Let $f(x) ∈ F[x] ⊂ K[x]$. Suppose that $f(x)$ is irreducible in $K[x]$.
$a)$ Prove that $f(x)$ is also irreducible in $F[x]$.
$b)$ Is it true that if $f(x)$ is irreducible in $F[x]$, then it is irreducible in $K[x]$, if not, give an example.
My attempt:
$a)$ Since $f(x)$ is irreducible over $K$, then $K[x]/(f(x))$ is a field (I have previously proven this).
But since $F ⊂ K$, then $F[x] ⊂ K[x]$ and thus, $F[x]/(f(x))⊂K[x]/(f(x))$.
Since $F[x]/(f(x))$ is a subfield, then it is a field, and so $f(x)$ is irreducible over $F$.
Is my attempt correct?
And I don't think this still holds if $f(x)$ is irreducible over $F$. Can someone please clarify part $(b)$ and give an example? Thank you
 A: Your proof for (a) isn't quite there, although the result is true. You correctly have an inclusion $F[x]/(f) \subseteq K[x]/(f)$ but $K[x]/(f)$ being a field does not a priori mean that $F[x]/(f)$ is a field. You have shown that it is a subring, but it is not true that all subrings of fields are field. For instance, $\mathbb Z \subseteq \mathbb Q$. This sort of thing cannot happen in this case, however, but you have not proven that it will not.
By the way, you can actually just stop here without doing all the work to show that $F[x]/(f)$ is a field. It is a subring of the field $K[x]/(f)$ so it is a domain. Thus, $(f) \subseteq F[x]$ is a prime ideal, so $f$ is prime. Hence, it is irreducible. I'd still like to show you that you can prove that $F[x]/(f)$ is a field, however.
The key is that $F[x]/(f)$ is not just some random subring of $K[x]/(f)$ - it contains $F$, i.e. it is an $F$-algebra. This will let us show that $F[x]/(f)$ must be a field. Indeed, we have inclusions $F \subseteq F[x]/(f) \subseteq K[x]/(f)$. In particular, we know that $F[x]/(f)$ is a domain. Furthermore, by polynomial long division, $F[x]/(f)$ is a vector space over $F$ of finite dimension. Indeed, let $n = \deg f$ and let $\overline g$ denote the image of $g \in F[x] \longrightarrow F[x]/(f)$. By polynomial long division, we can write $g = qf + r$ where $\deg r < n$ (we take the convention that $\deg 0 = -\infty$). Hence, $\overline{g} = \overline{r}$ and $\overline{r} = \sum_{i=0}^n r_i \overline{x^i}$. Thus, $1, \overline{x}, \dots, \overline{x^n}$ spans $F[x]/(f)$ as an $F$ vector space.
This puts us in the following general situation: let $F$ be a field and $D \supseteq F$ be an $F$ algebra that is also a domain. Suppose additionally that $D$ is a finite dimensional vector space over $F$. Then $D$ is a field. Indeed, take any $0 \neq d \in D$ and consider the map $D \longrightarrow D$ via $a \mapsto ad$. This map is injective as $D$ is a domain. Furthermore, it is $F$-linear: for $\lambda \in F$, $(\lambda a) \mapsto (\lambda a) d = \lambda (ad)$. Thus, this map $D \longrightarrow D$ is an injective linear transformation of finite dimensional $F$-vector spaces. By linear algebra, it is a bijection. In particular, it is onto so $1$ is in the image. That is, there exists some $a \in D$ such that $ad = 1$. Hence, $d$ is a unit in $D$ so $D$ is a field.
For part (b), here's an explicit counterexample. Let $F = \mathbb R$ and $K = \mathbb C$. Then the polynomial $x^2 + 1$ is irreducible in $\mathbb R[x]$, as if it were reducible it'd split into linear factors but you can show that this has no real roots. However, it is reducible in $\mathbb C[x]$ as $(x - i)(x + i)$. More generally, if $f \in F[x]$ is irreducible then $F \subseteq F[x]/(f)$ is a field and $f$ is reducible in $F[x]/(f)$. In fact, it has a root $\overline{x}$.
