On extensions of fields and embeddings

Question

This may be a simple doubt. As far as I know, when we say $K$ is a subfield of $L$, we mean that $K$ can be embedded in $L$ (through the embedding say $\phi$). Then, when I say, 'let $\alpha$ be in $L$ but not in $K$', I mean, $\alpha$ is not in the image of $\phi$.

Firstly, is it what the general understanding? Secondly, doesn't then the non-belongingness of $\alpha$ in $K$ depend on the embedding we choose? Can we have more than one injective maps from $K$ to $L$ in any circumstance? If so, why are we not stressing on the embedding we choose in the given situation of extensions?

For instance, suppose we consider the polynomial $p(x)=x^3-2$ over $\mathbb{Q}$. Let $\mathbb{Q}(a), \mathbb{Q}(b)$ and $\mathbb{Q}(c)$ be the three extensions obtained by adjoining the three roots $a(=\sqrt[3]{2}), b$ and $c$ of $p(x)$. Now, can't I embed $\mathbb{Q}(b)$ in $\mathbb{C}$ in two different ways? Firstly, we can directly treat $\mathbb{Q}(b)$ a subfield of $\mathbb{C}$ through the usual embedding. Secondly, I can treat $\mathbb{Q}(b)$ isomorphic to $\mathbb{Q}(a)$ and then I would get a different embedding. Now in the first embedding I see that $b$ is in the image space while in the second it is not.

Am I thinking in right direction?

Answer 1

As far as I know, when we say $K$ is a subfield of $L$, we mean that $K$ can be embedded in $L$ (through the embedding say $\phi$).

People play fast and loose with this terminology, but strictly speaking, no, this is not what this means. "$K$ is a subfield of $L$" means that $K$ is a subset of $L$ which is closed under the field operations. This is equivalent to saying that we have fixed an embedding of $K$ into $L$ but the point of insisting on $K$ being a literal subset is that it's totally unambiguous what it means for an element of $L$ to not be in $K$.

Secondly, doesn't then the non-belongingness of $\alpha$ in $K$ depend on the embedding we choose?

It would in general, but we've fixed it.

People are sometimes sloppy about this sort of thing, but one reason it's possible to get away with this is the following:

Proposition: If $K$ is a Galois extension of $\mathbb{Q}$, then if an embedding $K \to L$ exists, its image is unique (although the embedding is not itself unique; different embeddings are related by the action of the Galois group $\text{Gal}(K/\mathbb{Q})$).

In this case, despite the fact that there are multiple embeddings, the meaning of "an element of $L$ not in $K$" is still unambiguous. For example there are two embeddings $\mathbb{Q}[x]/(x^2 + 1) \to \mathbb{C}$ but they have the same image, namely the subfield $\mathbb{Q}(i)$ generated by $i$ (or $-i$). Your example of $\mathbb{Q}[x]/(x^3 - 2)$ is ambiguous because it's not Galois.