HW11

Let $L$ and $M$ be intermediate fields in the extension $K\subset F$.

• (a) $[LM:K]$ is finite if and only if $[L:K]$ and $[M:K]$ are finite.
• (b) If $[LM:K]$ is finite, then $[L:K]$ and $[M:K]$ divide $[LM:K]$ and $[LM:K]\le [L:K][M:K]$.
• (c) If $[L:K]$ and $[M:K]$ are finite and relatively prime, then $[LM:K]=[L:K][M:K]$.
• (d) If $L$ and $M$ are algebraic over $K$, then so is $LM$.

\begin{proof} (a) Assume that $[LM:K]$ is finite, then $[LM:K]=[LM:L][L:K]=[LM:M][M:K]$ yields that $[L:K]$ and $[M:K]$ are finite. Now assume that $[L:K]$ and $[M:K]$ are finite. Then $L$ and $M$ are finitely generated extension and they are algebraic extension. It follows that the subfield of $F$ generated by $L\cup M$ is also finitely generated and algebraic extension.Therefore, $[LM:K]$ is finite.

(b) Since $[LM:K]=[LM:L][L:K]=[LM:M][M:K]$, $[L:K]$ and $[M:K]$ divide $[LM:K]$. Suppose that $[L:K]=n$ and $[M:K]=m$, then $L={\oplus }_{i=1}^{s}K{\ell }_i$ with ${\ell }_i\in L$ and $M={\oplus }_{j=1}^{t}K{m}_j$ with $m_j\in M$. Note that $\{{\ell }_i{m}_j:1⩽i⩽s,1⩽j⩽t\}$ contains a basis of $LM$ over $K$, then $[LM:K]⩽[L:K][M:K]$.

(c) Since $([L:K],[M:K])=1$ and they divide $[LM:K]$, we have $[L:K][M:K]\mid [LM:K]$ and so $[L:K][M:K]⩽[LM:K]$. On the other hand, $[LM:K]⩽[L:K][M:K]$ by (b). Therefore, $[LM:K]=[L:K][M:K]$.

(d) Define algebraic closure of $K$ as $\overline{K}$. As $L$ and $M$ are algebraic over $K$, $\overline{K}$ contains $L$ and $M$. It follows that $\overline{K}$ contains $LM$ and so $LM$ is algebraic over $K$. \end{proof}

Remark. There is some question for d), see here. For d), we can use a) to prove it. See here.

Let $k$ be a field, let $A=k[x]$. Let $M=Ae\oplus Af$. Let $M^{\prime }\subset M$ be the submodule generated by $x(x+1)e$ and $\left(x^2-1\right)f$. Find a nice basis of $M^{\prime }$ as in the theorem about modules over PID.

\begin{proof} Let $e^{\prime }=xe+(x-1)f$, and let $f^{\prime }=e+f$. Note that

$$\left[\begin{array}{c}{e}^{\prime }\\ f^{\prime }\end{array}\right]=\left[\begin{array}{cc}x& x-1\\ 1& 1\end{array}\right]\left[\begin{array}{c}e\\ f\end{array}\right]$$

yields that $\{e^{\prime },f^{\prime }\}$ is a basis of $M$. Moreover, $M^{\prime }=⟨(x+1)e^{\prime },(x^3-x)f^{\prime }⟩$ as

$$\left[\begin{array}{cc}(x^2+x)e& (x^2-1)f\end{array}\right]\left[\begin{array}{cc}1& x-1\\ 1& x\end{array}\right]=\left[\begin{array}{cc}{e}^{\prime }& f^{\prime }\end{array}\right]$$

and $\left[\begin{array}{cc}1& x-1\\ 1& x\end{array}\right]$ is invertible. Hence, $\{(x+1)e^{\prime },(x^3-x)f^{\prime }\}$ is what we desired. \end{proof}

If $f\in K[x]$ has degree $n$ and $F$ is a splitting field of $f$ over $K$, then $[F:K]$ divides $n!$. (Be careful: $f(x)$ might be reducible)

\begin{proof} We prove it by induction. When $n=2$ and $f$ is reducible, $F=K$ and $[F:K]=1$ divides $2!$. If $f$ is irreducible, then $F\simeq K[x]/(f(x))$ and $[F:K]=2$ divides $2!$. Assume that the statement holds when $n⩽k-1$ and now consider the case of $n=k$.

If $f$ spilts in $K$, then $F=K$ and $[F:K]=1$ divides $n!$.

If $f$ is irreducible over $K$, then $f(x)$ has a root $u$ in $K_1=K[x]/(f(x))$ and $f(x)=(x-u)f_1(x)$ with $f_1\in K_1[x]$. By induction hypothesis, $[F:K_1]\mid (n-1)!$. Since $[K_1:K]=n$, we have $[F:K]=[F:K_1][K_1:K]$ divides $n!$.

If $f$ is reducible over $K$, then $f=g_1{g}_2$ with $n_i:=\mathrm{deg}{g}_i<\mathrm{deg}f$. Assume that $K_1$ is a splitting field of $g_1$ over $K$ and $K_2$ is a splitting field of $g_2$ over $K_1$, then $f$ spilts over $K_2$. By induction hypothesis, $[K_2:K_1]\mid n_2!$ and $[K_1:K]\mid n_1!$. Note that $\left(\genfrac{}{}{0ex}{}{n}{n_1}\right)=n!/n_1!n_2!$, and it follows that $n_1!n_2!\mid n!$. Hence, $[K_2:K]=[K_2:K_1][K_1:K]\mid n!$. Since $F$ is a subfield of $K_2$, we also have $[F:K]\mid n!$. Now we finish the proof. \end{proof}

If $[F:K]=2$, then $F$ is normal over $K$.

\begin{proof} For any $f(T)\in K[T]$ which has one root in $F$, we aim to show $f$ spilts in $F$. If $f$ is reducible in $K$, then $f$ spilts in $K$ and so for $F$. Otherwise assume that $f$ is irreducible, then there exists $u\in F\setminus K$ such that $f(u)=0$. Suppose that $f(T)=T^2+aT+b$. Define $v:=b/u\in F$, then note that

$$f(v)=v^2+av+b=\frac{b}{u^2}(b+au+u^2)=0.$$

Hence $v$ is a root of $f$ and $v\in F$. Therefore, $f(T)=(T-u)(T-v)$ spilts in $F$. By the arbitrary of $f$, $F$ is normal over $K$. \end{proof}