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

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 $\alpha \in E$.

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.

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 $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?