Why Automorphisms in Galois theory (why not Homomorphisms to itself)?

Question

I am wondering, why are Automorphism needed in Galois theory, why is it not enough that we have group homomorhisms $f$, that obey $$f(a+b)=f(a)+f(b)$$ and $$f(a\cdot b)=f(a)\cdot f(b)$$ where both $x$ and $f(x)$ are in the same field, and $f(0)=0$ and $f(1)=1$.

In other words, why do the mappings in question, $f$, have to be bijective, alongside obeying the above two properties, ie. why do they have to be isomorphisms, not just homomorphisms.

I have read this answer, but I still wonder about this.

Answer 1

As Arthur said, automorphisms form a group but endomorphisms won't. This is used heavily in Galois theory.

Actually an endomorphism of fields is quite often necessarily also surjective (it is always injective, because fields don't have non-trivial ideals, so kernel is trivial). Listing a few cases:

• If $K$ is finite dimensional over its prime field $F$ (either $\Bbb{Q}$ or $\Bbb{F}_p$ for some prime $p$), then $\sigma:K\to K$ is necessarily onto. This is a consequence of rank-nullity. The kernel is trivial, so $\dim_F\sigma(K)=\dim_F(K)$.
• If $K$ is algebraic over $F$ then, again, $\sigma:F\to F$ is always surjective. This is because each and every element of $K$ has only finitely many $F$-conjugates in $K$. Clearly $\sigma$ must stabilize any such set of conjugates, so injectivity (on a finite set!) again implies surjectivity.

OTOH the presence of transcendental elements gives us elbow room.

• The field $\Bbb{Q}(\pi)$ is isomorphic to its proper subfield $\Bbb{Q}(\pi^2)$, so here we have endomorphisms.
• Actually by Lüroth's theorem any intermediate field between $K$ and $K(x)$, $x$ transcental over $K$ is isomorphic to $K(x)$.
• The field $\Bbb{C}$ is isomorphic to some of its carefully selected proper subfields.

Answer 2

Because a set of automorphisms with composition forms a group, and groups are nice. That being said, any ring homomorphism $f:K\to R$ with $f(1)\neq0$ and $K$ a field must be injective, so it's not too restrictive to require it to be surjective as well.

Answer 3

In my experience, the question "What isomorphisms $F \to F$ are there?" is one that actually comes up in mathematics.

The question "What morphisms $F \to F$ are there?", however, only tends to come up when the question is "what morphisms $F \to E$ are there?" and, by happenstance, $F$ and $E$ happen to be the same field; there really isn't interest in the fact the set of morphisms being a monoid.