$\mathbb{C}$ is not the splitting field of any polynomial over $\mathbb{Q}$ (without cardinality)

Question

I came across this exercise in Gallian's Contemporary Abstract Algebra. Using a cardinality argument, the splitting field of polynomial over $\mathbb{Q}$ would form a finite-dimensional rational vector space, while $\mathbb{C}$ is an infinite-dimensional rational vector space. Is there a different way to go about this problem?

Answer 1

The existence of a transcendental number provides such a proof.

Answer 2

Maybe this is also interesting to you: Let $p_i$ for $i \in \mathbb{N}$ be pairwise different prime numbers. Then the set $\{log(p_i): i \in \mathbb{N}\}$ is linearly independent over the rationals. So not even the real numbers can be a finite dimensional vector space over the rationals. In particular the complex number can't be the splitting field of a single polynomial.

Answer 3

Another (slightly different) way to attack this problem: suppose $\;\Bbb C\;$ is the splitting field of $\;p(x)\in\Bbb Q[x]\;$, and let $\;w_1,...,w_n\;$ be the roots of $\;p(x)\;$ .

Then $\;p(x)\;$ already splits in $\;\Bbb Q(w_1,...,w_n)\subsetneq\Bbb C\;$ . The sharp contention follows either by dimension arguments or else because, for example,

$$\pi\in\Bbb C\setminus\Bbb Q(w_1,...,w_n)\;$$