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Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. Let $K = \mathbb{Q}(\sqrt{d})$ and $\overline{K}$ defined to be the algebraic closure of $K$. Is it true that $\overline{K} \cong \overline{\mathbb{Q}}$?

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"the"? A field has many algebraic closures. –  Chris Eagle Feb 10 '13 at 0:10
    
Oops, sorry! I did actually mean isomorphic. Thanks for pointing it out. –  user60194 Feb 10 '13 at 0:13
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The question is not whether the two fields are isomorphic but whether the algebraic closure of $\mathbb{Q}$ in $\bar{K}$ is all of $\bar{K}$. (To see that these two questions are not equivalent consider the embedding of $\mathbb{C}$ into $\mathbb{C}(t)$. As it turns out, the algebraic closure of $\mathbb{C}(t)$ is isomorphic to $\mathbb{C}$, but the embedding of $\mathbb{C}$ into it is not an isomorphism.) In general, the question people want to ask is usually not "are $A$ and $B$ isomorphic?" but "is this particular map I'm thinking of an isomorphism between $A$ and $B$?". –  Qiaochu Yuan Feb 10 '13 at 0:18
    
@Qiaochu: To add rigor while maintaining the spirit of the question, it might be better to phrase it as asking if the field extension $\overline{K}/\mathbb{Q}$ is isomorphic to $\overline{\mathbb{Q}}/\mathbb{Q}$. –  Hurkyl Feb 10 '13 at 1:18
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@Hurkyl: But that doesn't change the question -- any isomorphism between fields of characteristic $0$ is an isomorphism over $\mathbb{Q}$. –  Pete L. Clark Feb 10 '13 at 3:06
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2 Answers

up vote 4 down vote accepted

Algebraic closures are only defined up to isomorphism, so it doesn't make sense to ask if $\overline K$ and $\overline{\Bbb{Q}}$ are equal. It does make sense to ask if they are isomorphic, which they are: $\overline K$ is an algebraic extension of $\Bbb Q$ (since it's an algebraic extension of an algebraic extension) and is algebraically closed, and hence is an algebraic closure of $\Bbb Q$.

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If $d \in \mathbb{Q}$, $K$ will be a subfield of $\overline{\mathbb{Q}}$, since $\sqrt{d} \in \overline{\mathbb{Q}}$. Since $\overline{\mathbb{Q}}$ is algebraically closed, it suffices to show that $\overline{\mathbb{Q}}$ is algebraic over $K$. But if $\alpha \in \overline{\mathbb{Q}}$, choose a polynomial in $\mathbb{Q}[x]$ with $\alpha$ as a root. But then this polynomial also has coefficients in $K$, and so $\alpha$ is algebraic over $K$, too. Hence $\overline{\mathbb{Q}}$ is an algebraically closed field that is algebraic over $K$. These are the defining properties of an algebraic closure, so $\overline{K} = \overline{Q}$. Up to isomorphism, of course.

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