Is it possible to get the same extension field E/F of F by using different homomorphisms?

Question

Is it possible that we have 2 homomorphisms $i_{1}:F\rightarrow E$, $i_{2}:F\rightarrow E$ from the original field to the same extension field $E$?

Answer 1

This seems straightforward, unless I am misunderstanding your question.

Let $\varphi$ be a non-trivial automorphism of $F$. Define $i_2 := i_1\circ \varphi$.

Answer 2

• If $F$ has nontrivial automorphisms, it is enough to compose with one of them. For instance $\mathbb{F}_4=\{0,1,x,x+1=x^2\}$ has an automorphism $\phi: \;a \mapsto a^2$ that fixes $\mathbb{F}_2$ inside it and permutes the other two elements. If you consider any extension of $\mathbb{F}_4$, for instance $\mathbb{F}_4[t]$ where $t$ is a root of $t^2 + t + x$ (or any other grade $2$ irreducible polynomial over $\mathbb{F}_4$, there are six), then $$\theta : \mathbb{F}_4 \to \mathbb{F}_4[t]$$ that consists of $0 \mapsto 0$, $1\mapsto 1$, $x\mapsto x$, $x+1 \mapsto x+1$ and $$\theta \circ \phi: \mathbb{F}_4 \to \mathbb{F}_4[t]$$ that consists of $0 \mapsto 0$, $1\mapsto 1$, $x\mapsto x+1$, $x+1 \mapsto x$

• Even if $F$ does not have nontrivial automorphisms, it is still possible. You just need an automorphism of $E$ that does not restrict to an automorphism of $F$. Consider the splitting field $E$ over $\mathbb{Q}$ of $x^3-2$, and consider the extension $$\mathbb{Q} \subset \mathbb{Q}[^3\sqrt{2}] \subset K$$ There is an automorphism of $E$ which does not fix $F=\mathbb{Q}[^3\sqrt{2}]$ inside $E$ (here you really need to specify what map initially realizes this "inside"), given by $^3\sqrt{2} \to \omega \; ^3\sqrt{2}$ where $\omega$ is a complex cube root of unity. Composing any map $F \to E$ with that one gives another, different homomorphism.