Is Hartshorne exercise II.2.15(b) correct as written? Here is the text of the exercise:

If $f:X \rightarrow Y$ is a morphism of schemes over $k$, and $P \in X$ is a point with residue field $k$, then $f(P) \in Y$ also has residue field $k$.

In the case where $X$ and $Y$ are spectra of fields, this reduces to:

If $K \subset L \subset M$ are fields and $K \cong M$, then $K \cong L$.

which I believe is false, or else we could derive, for example, $\mathbb C \cong \mathbb C (x)$.
Am I mistaken?
 A: The problem here is that you have forgotten the condition that $X\to Y$ be "a morphism of schemes over $k$". This condition applied in your case means that the isomorphism $K\to M$ is a morphism of $K$-algebras. Once we enforce this, the sequence of inclusions $K\subset L\subset M$ is an inclusion of $K$-vector spaces, so $K=L=M$ by dimension considerations. 

Edit: This question ends up hinging on different interpretations of what exactly the statement "$P\in X$ is a point with residue field $k$" means. $k(P)$ is equipped with the structure of a $k$-algebra from the fact that $X$ is a scheme over $k$, and the intended statement here is that $k(P)$ with the $k$-algebra structure coming from $X$ being a scheme over $k$ is isomorphic to $k$ with the trivial $k$-algebra structure as a $k$-algebra, which is a more restrictive condition than just $k(P)\cong k$ as fields, as you have noticed. If one enforces this intended statement, then all opportunity for funny business is eliminated and the original answer, preserved above, is sufficient.
A: Here is where I believe I went wrong.
The correctness of the statement depends on what is meant by "with residue field $k$".
If "with residue field $k$" means "with a residue field isomorphic, as a field, to $k$", then the statement is incorrect. In this interpretation, the case when $X$ and $Y$ are spectra a fields reduces to:

If $K \subset L \subset M$ are fields and $K \cong M$, then $K \cong L$.

and this is false.
However, if "with residue field $k$" means "with residue field, viewed as a field over $k$, equal to $k$", then the statement is correct. In this interpretation, the case when $X$ and $Y$ are spectra of fields reduces to:

If $K \subset L \subset M$ are fields and $K = M$, then $K = L$.

which is trivially true.
But the second interpretation is the more natural in this context. The statement is about a morphism over $k$, so we should interpret the entire statement as being "over $k$". 
