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?



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 $$\{\alpha^k\beta^l \mid 0\le k<n,\, 0\le l<m\}$$ generates $\mathbb{Q}(\alpha,\beta)$; that is, $\mathbb{Q}(\alpha,\beta)$ is generated by a finite set.

Hanul Jeon
Hanul Jeon
October 20, 2019 05:17 AM

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