- #1
mathmari
Gold Member
MHB
- 5,049
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Hey! 
I am looking at the following exercise:
Let $C$ be an algebraic closure of $F$, let $f\in F[x]$ be irreducible and let $a,b\in C$ be roots of $f$.
Applying the theorem:
"If $E$ is an algebraic extension of $F$, $C$ is an algebraic closure of $F$, and $i$ is an embedding (that is, a monomorphism) of $F$ into $C$, then $i$ can be extended to an embedding of $E$ into $C$. "
show that there is a $F$-monomorphism $\tau :C\hookrightarrow C$, that maps $a$ to $b$ and how then that $\tau$ is onto, so it is an automorphism of $C$. I have done the following:
We have that $C$ is an algebraic closure of $F$ and $a$ and $b$ are elements of $C$, so $C$ is also an algebraic closure of $F[a]$ and $F$.
Then we have the algebraic extensions $C\leq F[a]$ and $C\leq F$.
Since $F[a]$ and $F$ are subfields of $C$, there are the field monomorphisms $\tau_{a}: F[a]\rightarrow C$ and $\tau_{b}: F\rightarrow C$.
We have that $f\in F[X]$ is irreducible, then there is a field that contains $F$, in which $f(X)$ has a root.
This field is $K=F[X]/\langle f(X)\rangle$.
We have that $K=F[a]$ and $K=F$.
So, we have that $F[X]/\langle f(X)\rangle=F[a]$ and $F[X]/\langle f(X)\rangle=F$.
Therefore, we have the field monomorpshims $\tau_{a}: F[X]/\langle f(X)\rangle\rightarrow C$ with $\tau_{a}(g(X))=a$ and $\tau_{b}: F[X]/\langle f(X)\rangle\rightarrow C$ with $\tau_{b}(g(X))=b$.
Applying the above theorem with $E:=C$ and $F:=F[X]/\langle f(X)\rangle$, we have the monomorphism $\tau : C\rightarrow C$ with $\tau \circ \tau_{a}=\tau_{b}$.
So, $\tau ( \tau_{a}(g(X)))=\tau_{b}(g(X)) \Rightarrow \tau ( a)=b$.
Is this correct? (Wondering) From $\tau:C\rightarrow C$ we have that $\tau (C)\subseteq C$.
We have that $F\leq C\Rightarrow \tau (F)\subseteq \tau (C)$.
Since $\tau$ is a $F$-monomoprhism, do we have that $\tau (F)=F$ ? (Wondering)
I am looking at the following exercise:
Let $C$ be an algebraic closure of $F$, let $f\in F[x]$ be irreducible and let $a,b\in C$ be roots of $f$.
Applying the theorem:
"If $E$ is an algebraic extension of $F$, $C$ is an algebraic closure of $F$, and $i$ is an embedding (that is, a monomorphism) of $F$ into $C$, then $i$ can be extended to an embedding of $E$ into $C$. "
show that there is a $F$-monomorphism $\tau :C\hookrightarrow C$, that maps $a$ to $b$ and how then that $\tau$ is onto, so it is an automorphism of $C$. I have done the following:
We have that $C$ is an algebraic closure of $F$ and $a$ and $b$ are elements of $C$, so $C$ is also an algebraic closure of $F[a]$ and $F$.
Then we have the algebraic extensions $C\leq F[a]$ and $C\leq F$.
Since $F[a]$ and $F$ are subfields of $C$, there are the field monomorphisms $\tau_{a}: F[a]\rightarrow C$ and $\tau_{b}: F\rightarrow C$.
We have that $f\in F[X]$ is irreducible, then there is a field that contains $F$, in which $f(X)$ has a root.
This field is $K=F[X]/\langle f(X)\rangle$.
We have that $K=F[a]$ and $K=F$.
So, we have that $F[X]/\langle f(X)\rangle=F[a]$ and $F[X]/\langle f(X)\rangle=F$.
Therefore, we have the field monomorpshims $\tau_{a}: F[X]/\langle f(X)\rangle\rightarrow C$ with $\tau_{a}(g(X))=a$ and $\tau_{b}: F[X]/\langle f(X)\rangle\rightarrow C$ with $\tau_{b}(g(X))=b$.
Applying the above theorem with $E:=C$ and $F:=F[X]/\langle f(X)\rangle$, we have the monomorphism $\tau : C\rightarrow C$ with $\tau \circ \tau_{a}=\tau_{b}$.
So, $\tau ( \tau_{a}(g(X)))=\tau_{b}(g(X)) \Rightarrow \tau ( a)=b$.
Is this correct? (Wondering) From $\tau:C\rightarrow C$ we have that $\tau (C)\subseteq C$.
We have that $F\leq C\Rightarrow \tau (F)\subseteq \tau (C)$.
Since $\tau$ is a $F$-monomoprhism, do we have that $\tau (F)=F$ ? (Wondering)