Why is it the smallest subfield containing F and $\alpha$?

Question

Please take a look at the sentence in red:

I understand that $\phi_\alpha[F[x]]$, is a subfield which contains $\alpha$, and F(we just need to evaluate $\phi_\alpha$ at the appropriate values). But why is it the smallest subfield containing these two?

In order to show this must we not assume that K is a subfield of E, and then show that $\alpha \in K, F \subset K$? Is this difficult to show? Can you please help me?

Answer 1

It's because if $K$ contains $\alpha$, it also must contain any power of $\alpha$, and any linear combination of these powers with coefficients in $F$. Isn't that the definition of the image of $F[x]$ under $\phi_\alpha$?

Answer 2

Not too difficult: let $\alpha \in K\subset E$ with $F\subset K$. Set $F(\alpha) = \phi_{\alpha}(F[x])$.

Then, since $F[x]$ is generated by $F$ and $x$, we see that $F(\alpha)$ is generated by $F$ and $\phi_{\alpha}(x) = \alpha$.

Since $F\subset K$ and $\alpha\in K$, then the field generated by $F$ and $\alpha$ is also in $K$. But this is just $F(\alpha)$, so $F(\alpha)\subset K$.

Answer 3

Suppose $K$ is any subfield of $E$ containing $\alpha$ and $F$. Let $f(x)$ be any element of $F[x]$, say:

$f(x) = a_0 + a_1x + \cdots + a_nx^n$.

Now $\phi_{\alpha}(f(x)) = a_0 + a_1\alpha + \cdots + a_n\alpha^n$.

Since $F \subset K$, we have that $a_0,\dots,a_n \in K$. Since $\alpha \in K$, we have that: $\alpha,\cdots,\alpha^n \in K$, by closure of multiplication in a field.

Also by closure of multiplication, we have that $a_k\alpha^k \in K$, for each $k = 0,1,\dots,n$.

Finally, by closure of addition, we have $\phi_{\alpha}(f(x)) \in K$, for any $f(x) \in F[x]$.

Now $\phi_{\alpha}$ is surjective onto its image, so any element of $\phi_{\alpha}(F[x])$ has just such a pre-image $f(x) \in F[x]$. Thus every element of $\phi_{\alpha}(F[x])$ lies within $K$. This shows that $\phi_{\alpha}(F[x])$ is indeed the smallest subfield of $E$ containing $F$ and $\alpha$, as any other such field contains it.