HW12

Show the following:

• (a) $\sqrt{3}\notin \mathbb{Q}[\sqrt{2}]$.
• (b) $\sqrt{5}\notin \mathbb{Q}[\sqrt{2},\sqrt{3}]$.

\begin{proof} (a) Since $[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]=\mathrm{deg}\mathrm{Irr}(\sqrt{2},\mathbb{Q})=2$, $\mathbb{Q}[\sqrt{2}]$ is a $\mathbb{Q}$-vector space with basis $\{1,\sqrt{2}\}$. If $\sqrt{3}\in \mathbb{Q}[\sqrt{2}]$, then there exists $a,b\in \mathbb{Q}$ such that $a+b\sqrt{2}=\sqrt{3}$, which is impossible because $a^2+2b^2+2ab\sqrt{2}\in \mathbb{Q}$ with $ab\ne 0$.

(b) Note that $[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}[\sqrt{2}]][\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]$. By (a) $\sqrt{3}\notin \mathbb{Q}[\sqrt{2}]$ and so $[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}[\sqrt{2}]]=\mathrm{deg}\mathrm{Irr}(\sqrt{3},\mathbb{Q}[\sqrt{2},\sqrt{3}])=2$. It follows that $[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}]=4$. Note that $\mathbb{Q}[\sqrt{2}+\sqrt{3}]=\mathbb{Q}[\sqrt{2},\sqrt{3}]$, then $\mathrm{Irr}(\sqrt{2}+\sqrt{3},\mathbb{Q})=(x^2-1{)}^2-8x^2$ and so $\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},11\sqrt{2}+9\sqrt{3}\}$ is a basis of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ as $\mathbb{Q}$ vector space.

Assume that $\sqrt{5}\in \mathbb{Q}[\sqrt{2},\sqrt{3}]$, then there exists $a,b,c,d\in \mathbb{Q}$ such that $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=\sqrt{5}$. It follows that $3cd\sqrt{2}+2bd\sqrt{3}+bc\sqrt{6}\in \mathbb{Q}$. Since $\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},11\sqrt{2}+9\sqrt{3}\}$ is a basis, $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is also a basis and $cd=bd=bc=0$. Thus, one of

$$a+b\sqrt{2}=\sqrt{5},a+c\sqrt{3}=\sqrt{5},a+d\sqrt{6}=\sqrt{5}$$

holds. However, none of them possible. Therefore, $\sqrt{5}\notin \mathbb{Q}[\sqrt{2},\sqrt{3}]$. \end{proof}

Let $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. Find $x\in K$ such that $K=\mathbb{Q}[x]$ (i.e., $x$ is a primitive element).

\begin{proof} Recall that $f(x)=\mathrm{Irr}(\sqrt{2}+\sqrt{3},\mathbb{Q})=(x^2-1{)}^2-8x^2$ and the roots of $f$ are $\pm \sqrt{2}\pm \sqrt{3}$. Also, the roots of the minimal polynomial of $\sqrt{5}$ are $\pm \sqrt{5}$. By the proof of the theorem of primitive element, since

$$\sqrt{2}+\sqrt{3}+\sqrt{5}\notin \{\frac{\sqrt{2}+\sqrt{3}\pm \sqrt{2}\pm \sqrt{3}}{2\sqrt{5}}\},$$

we have $K=\mathbb{Q}[\sqrt{2},\sqrt{3}][\sqrt{5}]=\mathbb{Q}[\sqrt{2}+\sqrt{3},\sqrt{5}]=\mathbb{Q}[\sqrt{2}+\sqrt{3}+\sqrt{5}]$. Hence, $x=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is what we desired. \end{proof}

Let $a,b,c\in \mathbb{Q}$ be 3 nonzero elements. Show that $\mathbb{Q}[a\sqrt{2}+b\sqrt{3}+c\sqrt{6}]$ is the same as $\mathbb{Q}[\sqrt{2},\sqrt{3}]$.

\begin{proof} Since $\sqrt{6}\in \mathbb{Q}[\sqrt{2},\sqrt{3}]$, then $F:=\mathbb{Q}[a\sqrt{2}+b\sqrt{3}+c\sqrt{6}]\subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}]$. On the other hand, as $a\sqrt{2}+b\sqrt{3}+c\sqrt{6}\in F$, $3bc\sqrt{2}+2ac\sqrt{3}+ab\sqrt{6}$ and $(6a{c}^2+3a{b}^2)\sqrt{2}+(6b{c}^2+2a^{2}b)\sqrt{3}+(2a^{2}c+3b^{2}c)\sqrt{6}$ are contained in $F$. Note that

$$\{(a,b,c),(3bc,2ac,ab),(6a{c}^2+3a{b}^2,6b{c}^2+2a^{2}b,2a^{2}c+3b^{2}c)\}$$

are linearly independent, then $\sqrt{2},\sqrt{3},\sqrt{6}\in F$. Therefore, $F=\mathbb{Q}[\sqrt{2},\sqrt{3}]$. \end{proof}

Let $\alpha$ be a real number such that ${\alpha }^4=5$.

• (a) Show that $\mathbb{Q}\left(i{\alpha }^2\right)$ is normal over $\mathbb{Q}$.
• (b) Show that $\mathbb{Q}(\alpha +i\alpha )$ is normal over $\mathbb{Q}\left(i{\alpha }^2\right)$.
• (c) Show that $\mathbb{Q}(\alpha +i\alpha )$ is not normal over $\mathbb{Q}$.

\begin{proof} (a) Note that $\mathrm{Irr}(i{\alpha }^2,\mathbb{Q})=x^2+5$ and the roots are $\pm i{\alpha }^2\in \mathbb{Q}(i{\alpha }^2)$. Thus $\mathbb{Q}(i{\alpha }^2)$ is a splitting field over $\mathbb{Q}$ and so it is normal.

(b) Note that $\mathrm{Irr}(\alpha +i\alpha ,\mathbb{Q}(i{\alpha }^2))=x^2-2i{\alpha }^2$ and the roots are $\pm (1+i)\alpha \subseteq \mathbb{Q}(\alpha +i\alpha )$. Thus $\mathbb{Q}(\alpha +i\alpha )$ is a splitting field over $\mathbb{Q}(i{\alpha }^2)$ and so it is normal.

(c) By (a) and (b) we have

$$[\mathbb{Q}(\alpha +i\alpha ):\mathbb{Q}]=[\mathbb{Q}(\alpha +i\alpha ):\mathbb{Q}(i{\alpha }^2)][\mathbb{Q}(i{\alpha }^2):\mathbb{Q}]=4.$$

It deduces that $f(x)=\mathrm{Irr}(\alpha +i\alpha ,\mathbb{Q})=x^4+20$. The roots of $f$ are $\{(1+i)\alpha ,(-1+i)\alpha ,(-1-i)\alpha ,(1-i)\alpha \}$.

If $\mathbb{Q}(\alpha +i\alpha )$ is normal over $\mathbb{Q}$, then $(-1+i)\alpha \in \mathbb{Q}(\alpha +i\alpha )$ and so $\alpha ,i\in \mathbb{Q}$. It follows that $\mathbb{Q}(\alpha +i\alpha )=\mathbb{Q}(i,\alpha )$ and $[\mathbb{Q}(i,\alpha ):\mathbb{Q}]=[\mathbb{Q}(i,\alpha ):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=8$, which is a contradiction. Therefore, $\mathbb{Q}(\alpha +i\alpha )$ is not normal over $\mathbb{Q}$. \end{proof}

Let $F/K$ be an extension, and let $E$ be an intermediate field.

• (a) If $u\in F$ is separable over $K$, then $u$ is separable over $E$.
• (b) If $F$ is separable over $K$, then $F$ is separable over $E$ and $E$ is separable over $K$.

\begin{proof} (a) Since $u\in F$ is separable over $K$, we have $\mathrm{Irr}(u,K)$ is separable, that is, $\mathrm{Irr}(u,K)={\prod }_{i=1}^n(T-u_i)$ inside $\overline{K}[T]$ and $u_i$ are distinct. Since $E\supseteq K$, there is $\mathrm{Irr}(u,K)={\prod }_{i=1}^n(T-u_i)$ inside $\overline{E}[T]$ and $u_i$ are distinct. Therefore, $u$ is separable over $E$.

(b) If $F$ is separable over $K$, then for any $u\in F$, $u$ is separable over $K$ and so separable over $E$ by (a). Hence, $F$ is separable over $E$. For any $u\in E$, since $E\subseteq F$, $u$ is also separable over $K$. Therefore, $E$ is separable over $K$. \end{proof}

Remark. Furthermore, if $F/E$ and $E/K$ are separable, then $F/K$ is separable. See here.