# Closure of Algebraic Closure

by james black   Last Updated October 20, 2019 05:20 AM - source

Prove that the algebraic closure of Q is closed under multiplication.

So I assume Q is the rational numbers here because this wouldn't apply to any field and to prove it is closed under multiplication then if x,y ∈ Q, then xy ∈ Q.

I understand the definition and existance of algebraic closure for every field but am confused as how to apply the definition on how to prove the problem.

edit: i assume i prove the algebraic closure of Q is algebraic numbers which is a field, so i have to prove two parts but how would i proceed for each of those parts?

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#### Answers 1

I assume that you know some basic facts on field theory. I will prove that a multiplication of two algebraic numbers is again an algebraic number.

Let $$\alpha$$, $$\beta$$ be algebraic numbers over $$\mathbb{Q}$$. If you can show $$[\mathbb{Q}(\alpha \beta):\mathbb{Q}]$$ is finite, then you get $$\alpha\beta$$ is algebraic over $$\mathbb{Q}$$. Observe that $$\mathbb{Q}(\alpha\beta)\subseteq \mathbb{Q}(\alpha,\beta)$$, so we will prove that $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$$ is finite instead.

Let $$n$$ and $$m$$ be a degree of an minimal polynomial of $$\alpha$$ and $$\beta$$ respectively. Then $$\{1,\alpha,\cdots, \alpha^{n-1}\}$$ and $$\{1,\beta,\cdots,\beta^{m-1}\}$$ are bases of $$\mathbb{Q}(\alpha)$$ and $$\mathbb{Q}(\beta)$$ over $$\mathbb{Q}$$ respectively. Now, you can see that the set $$\{\alpha^k\beta^l \mid 0\le k generates $$\mathbb{Q}(\alpha,\beta)$$; that is, $$\mathbb{Q}(\alpha,\beta)$$ is generated by a finite set.

Hanul Jeon
October 20, 2019 05:17 AM