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Show $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a simple extension field of $\mathbb{Q}$.

Since $\sqrt{2},\sqrt{3}\in \mathbb{Q}(\sqrt{2},\sqrt{3})$, and $\sqrt{2}\sqrt{3}=\sqrt{6}\in \mathbb{Q}(\sqrt{2},\sqrt{3})$, so $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{6})$, hence $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a simple extension field of $\mathbb{Q}$.

Does my argument right? The only information I have is : A field extension $E$ of $F$ is called a simple extension if $E = F(\alpha)$ for some $\alpha \in E$.

Simple
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  • The answers given are correct but there is the primitive element theorem which states that for any field $F$ of characteristic $0$ and any finite extension $E/F$ of $F$ you can find $\alpha\in E$ such that $E=F(\alpha)$. Hence any finite extension over $\mathbb{Q}$ is simple. – Clément Guérin Jun 04 '15 at 06:33

2 Answers2

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Definition: $\def\Q{{\Bbb Q}}\Q(\alpha_1,\alpha_2,\ldots)$ is the smallest field containing $\Q$ and $\alpha_1,\alpha_2,\ldots\,$.

Lemma: $\Q(\sqrt2,\sqrt3)=\Q(\sqrt2+\sqrt3)$.

Proof. By definition, LHS is a field which contains $\sqrt2$ and $\sqrt3$. Since a field is closed under addition, LHS is a field containing $\Q$ and $\sqrt2+\sqrt3$. By definition, RHS is the smallest such field, so RHS${}\subseteq{}$LHS.

By definition, RHS is a field containing $\Q$ and $\alpha=\sqrt2+\sqrt3$. By the closure laws, RHS also contains the following:

  • $\alpha^3=11\sqrt2+9\sqrt3$;
  • $\alpha^3-9\alpha=2\sqrt2$;
  • $\frac12(\alpha^3-9\alpha)=\sqrt2$;
  • $\alpha-\sqrt2=\sqrt3$.

That is, RHS is a field containing $\Q$ and $\sqrt2$ and $\sqrt3$, so RHS${}\supseteq{}$LHS.

Hence $\Q(\sqrt2,\sqrt3)=\Q(\sqrt2+\sqrt3)$, and this is a simple extension.

David
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  • Therefore $ \mathbb Q( \sqrt2, \sqrt3): \mathbb Q$ is simple. (I think this is the missing conclusion of the proof, isn't? ) – Aaron Martinez May 17 '17 at 05:19
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No, your argument is not correct: just because $\sqrt{6}$ is an element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, does not mean that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{6})$. In fact $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4\quad\qquad [\mathbb{Q}(\sqrt{6}):\mathbb{Q}]=2$$ However, you will find this thread relevant: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

Community
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Zev Chonoles
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  • $\sqrt{2}+\sqrt{3}\in \mathbb{Q}(\sqrt{2},\sqrt{3})$, I may need to write a argument for $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, right? – Simple Jun 04 '15 at 06:25
  • Indeed, you will need an argument. After all, $0$ is an element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, that doesn't mean $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(0)=\mathbb{Q}$. – Zev Chonoles Jun 04 '15 at 06:26
  • @Simple since $\sqrt2+\sqrt3\in \Bbb Q(\sqrt2,\sqrt3)$, we have $\Bbb Q(\sqrt2+\sqrt3)\subseteq \Bbb Q(\sqrt2,\sqrt3)$. You also need to prove inclusion the other way, either by constructing $\sqrt2$ and $\sqrt3$ inside $\Bbb Q(\sqrt2+\sqrt3)$, or by a more abstract argument, like extension degrees. – Arthur Jun 04 '15 at 06:31
  • Do I need show $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is a field? – Simple Jun 04 '15 at 06:51