$(A)$ $L:K$ is normal
$(B)$ if $j$ is any monomorphism from $L$ to $\overline{L}$ which fixes $K$ then $j(L) \subseteq L$
$(C)$ if $j$ is any monomorphism from $L$ to $\overline{L}$ which fixes $K$ then $j(L) = L$
I'm looking at this, and I'm unsure about on how to do prove everything. I've attempted $(A) \implies (B)$ but I'm still unsure if the argument is right because I think I'm missing a few things in it.
But this is what I have so far:
$(A) \implies (B)$:
Suppose that $L:K$ is normal and that we have $j: L \rightarrow \overline{L}$ is a monomorphism which fixes $K$.
Suppose that $\alpha \in L$ and let $m$ be the minimal polynomial for $\alpha$ over $K$, this means that $m(j(\alpha)) = j(m(\alpha)) = 0$ which means that $j(\alpha)$ is a root of $m$. As $m$ splits over $L$, $j(\alpha) \in L$ which means $j(L) \subseteq L$
That's all I have so far, not sure how to... start the others. Or if this is correct either. Any help would be nice!
$\;(B)\implies (C)\;$ . Let $\;a\in L\;$ and let $\; a=a_1,a_2,...,a_n\;$ be all its conjugates (i.e., all the roots of the minimal polynomial $\;m_a(x)\;$ of $\;a\;$ over $\;K\;$ )
We know that for any $\;1\le i\le n\;$ exists $\;1\le k(i)\le n\; $ .s.t $\;j(a_i)=a_{k(i)}\;$, and since $\;j\;$ is $\;1-1\;$ we get that $\;\{a_1,...,a_n\}=\{j(a_1),...,j(a_n)\}\;$ , which means that $\;a=j(a_r)\;$ , for some $\;1\le r\le n\; \implies L\subset j(L)\;$ and we're done.
$\;(C)\implies (A)\;$ . Let $\;m(x)\in K[x]\;$ be irreducible with a root $\;a\in L\;$, and let us look at $\;K(a)/K\;$ . Let $\;b\;$ be any other root of $\;m\;$ in $\;\overline L\;$ .
There exists an isomorphism $\;K(a)\to K(b)\subset \overline L\;,\;\;f(a)\mapsto f(b)\;,\;\;\forall\,f(x)\in K[x]\;$ , which can be extended to a monomorphism $\;j:L\to \overline L\;$ . By assumption, $\;j(L)=L\;$ and thus $\;b\in L\implies m(x)\;$ splits in $\;L\;$ .