There always exists a subfield of $\mathbb C$ which is a splitting field for $f(x)$ $\in$ $Q[X]$?

Question

So I've been studying field theory on my own, and I just started learning about splitting fields. Based on my understandings if a polynomial, $f(x)$ $\in$ $Q[X]$, then there should be always a subfield of $\mathbb C$ which is a splitting field for that polynomial. Is this true? Thanks in advance.

Answer 1

It is true, and I'll provide a way of explicitly constructing the splitting field. For convenience, let's consider irreducible polynomials over $\mathbb{Q}$, and this will extend easily to reducible ones.

First, it is a theorem that, in general, if $f \in F[x]$ is irreducible, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$ where $\alpha$ is a root of $f$.

We can prove this with the isomorphism theorems. Consider the evaluation homomorphism $ev_{\alpha}:F[x] \rightarrow F[\alpha]$ defined as follows: $g(x) \mapsto g(\alpha)$. You'll want to prove that this is indeed a homomorphism.

Now certainly $\ker(\phi) = \langle f(x) \rangle$. This is because $F[x]$ is a PID and $f(x)$ is the minimal polynomial over $F$ with $\alpha$ as a root. Finally, we know that $ev_\alpha$ maps surjectively to $F[\alpha]$.

Hence, by the isomorphism theorems, we have $F[x]/\ker(\phi) \cong Im(\phi) \Longrightarrow F[x]/\langle f(x) \rangle \cong F[\alpha]$.

---

Of course, in this scenario, $F = \mathbb{Q}$. At this point, if our irreducible $f$ splits completely in $\mathbb{Q}[\alpha]$, then we're finished. If not, then say it splits as $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ are irreducible over the extension field. We can further extend $\mathbb{Q}[\alpha]$ to contain even more roots of $f$: $\mathbb{Q}[\alpha]/\langle g(x) \rangle \cong \mathbb{Q}[\alpha, \beta]$ where $\beta$ is a root of $g$. We can continue this process until $\mathbb{Q}[\alpha, \beta, \gamma, \cdots]$ contains every root of $f$.

Since $f$ has, at most, $\deg(f)$ roots, then we need only iterate this process at most $\deg(f)$ times. It will actually be less than that, but the point is that the process does terminate, and the splitting field will be of finite degree over $\mathbb{Q}$.

Of course, this is a rough sketch, but hopefully you can see the gist of where this is going and clean it up a bit on your own.

---

Footnote: The splitting field is always guaranteed to be a proper subfield of $\mathbb{C}$ since, for example, $\pi$ is not algebraic over $\mathbb{Q}$.

Answer 2

Yes. Every polynomial in $\Bbb C[x]$ has a root in $\Bbb C$ (fundamental theorem of algebra), which means that every polynomial has a linear factor; after dividing we get another polynomial in $\Bbb C[x]$ of smaller degree, which also has a root, which means we get another linear factor we can divide by, and we can continue this process until we're left with $1$, and when we collect all of the linear factors we've gathered (possibly with multiplicity) we have a complete factorization (over $\Bbb C$) of the original polynomial. Every polynomial in $\Bbb Q[x]$ is in also a polynomial in $\Bbb C[x]$.

Indeed, every field $k$ has an algebraic closure $K$. Every polynomial in $K[x]$ factors completely over the field $K$. Splitting fields (which algebraic closures are a particular example of) can be explicitly constructed by iteratively adjoining roots of irreducible factors over higher and higher extensions, and this process might need to be done transfinitely using Zorn's lemma; this should be covered in any text on field theory. (Another way I never see discussed anywhere is forming a splitting ring for polynomials using Vieta's formulas and then quotienting by a maximal ideal.)