Why is $F$-homomorphism so important in field theory?

Question

In group and ring theory we work with homomorphisms between groups and rings which preserves the structure.But in field theory,if $K,L$ are two extensions of $F$ then we work with $F$-homomorphisms (field homomorphism fixing $F$)between those fields instead of just taking field homomorphisms.I want to know why these $F$-homomorphisms are given so importance instead of just homomorphisms?I think it is so because it actually is a vector space homomorphism between $K$ and $L$ regarded as a vector space over $F$.

Answer 1

Remember that a homomorphism is supposed to preserve all of the relevant "structure" present in an algebraic gadget. So when you say that $K$ and $L$ are two extensions of $F$, you're implicitly answering your own question:

A "homomorphism of fields" between $K$ and $L$ only needs to preserve the field-theoretic structure, but if we're viewing $K$ and $L$ as extensions of $F$, then a "homomorphism" should preserve not only the field-theoretic structure, but also the "extension of $F$" structure. As you've noticed, this is related to the fact that $K$ and $L$ are additionally $F$-vector spaces, so a homomorphism $\varphi$ should guarantee that $\varphi(f \cdot k) = f \cdot \varphi(k)$ for $f \in F$ and $k \in K$. Of course, when $k=1$ this says that $f = \varphi(f)$ for each $f$, so that $F$ is fixed by $\varphi$.

Now, it's a perfectly reasonable follow up question as to why it's useful to study extensions of $F$. One algebraic reason is that extensions of $F$ are the natural home for solutions to equations we can write down in $F$, but which we can't solve in $F$ (think of $x^2 - 2 = 0$ as an equation over $\mathbb{Q}$. The natural solutions to this equation live in $\mathbb{Q}(\sqrt{2})$, an extension). Unsurprisingly, since we want to consider solutions to equations, it would be extremely annoying if, after appplying $\varphi$, the elements of $F$ were permuted! In that case, one equation might turn into another equation after applying $\varphi$, which gets in the way of comparing solutions.

There's a geometric reason as well, but it takes a bit of algebraic geometry to really understand. The gist is that if we work with any ring extensions of $F$, rather than just field extensions, then the $F$-homomorphisms admit a much more natural geometric interpretation than generic homomorphisms. Indeed, in the classical case of affine varieties over $\mathbb{C}$, a map in the geometric sense is exactly a homomorphism between the coordinate rings that fixes $\mathbb{C}$. While we can make sense geometrically of general ring homomorphisms, they tend to be much worse behaved than the homs which fix some shared base field, and behave in ways that are less geometrically natural.

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I hope this helps ^_^