Field homomorphism is onto

Question

Let $F$ be a field and $A:F \to F$ be a homomorphism. Show that $A$ is onto.

Abstract algebra proof. This question very important to me.

Answer 1

Let $F=\Bbb Q(X_1,X_2,\ldots)$ be the field of rational functions in countably many variables $X_k$, $k\in\Bbb N$. Then $X_k\mapsto X_{k+1}$ induces a field homomoprhism with $X_1$ not in its image.

Answer 2

That is not true. If we define the kernel of the homomorphism $A$ as:

$$\mathrm{Ker}(A):= \{x \in F : A(x) = 0_{\text F} \}\subseteq F $$ we know two things:

• $\mathrm{Ker}(A)$ is an ideal of $F$;
• The ideals of a field are only two: the null one $\{0_{\text F} \}$ and F.

So there are two possibilities:

1. $\mathrm{Ker}(A) = \{0_{\text F} \}$ ;
2. $\mathrm{Ker}(A) = F$.

Since $A$ is an homomorphism, then $A(1_{\text F} )=1_{\text F} \neq 0_{\text F} $ and therefore $1_{\text F} \notin \mathrm{Ker}(A)$.

It follows that $\mathrm{Ker}(A) = \{0_{\text F} \}$, so $A$ is injective. Note that, from Pigeonhole principle, we get that

If $R$ is a set is finite then every function $f \colon R \to R$ is onto if and only if is injective if and only if is surjective.

So we get this proposition:

If $F$ is a finite field and $A \colon F \to F$ is a homomorphism, then $A$ is onto.

But in infinite fields there are examples of injective homomorphisms that are not surjective, as Hagen von Eitzen showed you.