Existence of extension field containing all roots of a polynomial $f(x) \in \mathbb{F}[x]$.

Question

This is not for homework, but I would just like a hint please. The question asks

If $f(x) \in \mathbb{F}[x]$ has degree $n$, show that there exists an extension field $\mathbb{E}$ of $\mathbb{F}$ such that $\mathbb{E}$ contains all the roots of $f(x)$.

I believe induction is the way to go with this problem. In the case of $n = 1$, the extension field $\mathbb{F}[x]/(f(x))$ of $\mathbb{F}$ contains the root of $f(x)$. Suppose the claim is true for $\deg (f(x)) < n$ for some $n \geq 2$.

Suppose $\deg (f(x)) = n$. I know there is an extension field $\mathbb{K}$ of $\mathbb{F}$ that contains at least one root, say $c_n$, of $f(x)$. Now I can write $f(x) = (x - c_n)g(x)$, where $\deg (g(x)) < n$. The induction hypothesis then implies there exists an extension field $\mathbb{E}$ of $\mathbb{F}$ such that $\mathbb{E}$ contains all the roots of $g(x)$. Now, all the roots of $g(x)$ are also roots of $f(x)$. The part that is confusing me is that I have these two fields $\mathbb{E}$ and $\mathbb{K}$ containing all the roots of $f(x)$. Can I combine them somehow to get just one field containing the roots of $f(x)$? What am I doing wrong?

In "the induction hypothesis" do the extension on $;\Bbb K;$ , not on $;\Bbb F;$! That's the whole trick here... — DonAntonio, Oct 26 '13 at 18:44
Do not forget that coefficients of $g(x)$ lie in that extension field $K$, not in $F$. So the induction hypothesis can only give you an extension field of $K$. — Jyrki Lahtonen, Oct 26 '13 at 18:50
@DonAntonio: Could you post that as an answer? We are trying to reduce the number of questions in "unanswered" tab. I will upvote! Thanks :) — Prism, Nov 09 '13 at 09:03

score 5 · Accepted Answer · answered Nov 09 '13 at 12:06

Your way is the way to go, only taking into account what's in the comments:

$$f(x)\in\Bbb F[x]\;\;\text{irreducible}\implies \alpha:=x+\langle f(x)\rangle\in\Bbb K:=\Bbb F[x]/\langle f(x)\rangle\;\text{is a root of}\;f(x)$$

so that $\;f(x)=(x-\alpha)g(x)\;$ in $\;\Bbb K[x]\;$ and $\;[\Bbb K:\Bbb F]=n:=\deg f(x)\;$ .

Repeat the above (inductively) for $\;g(x)\;$ in $\;\Bbb K[x]\;$ (or one of its irreducible factors in case it is not irreducible over $\;\Bbb K\;$) and voilá, since $\;\deg g(x)<n\;$ .

Existence of extension field containing all roots of a polynomial $f(x) \in \mathbb{F}[x]$.

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