Let $k$ a field, $k'$ a subfield, and $A$ any associative $k$-algebra. Can a quotient of $A$ ever yield $k'$?

Question

I am trying to learn some basic Scheme theory out of Eisenbud's book "Schemes: the Language of Modern Algebraic Geometry." I'm trying to understand how elements of a ring can be treated as functions over spec in the classical sense. Of course, the title is a little broad for this interpretation. However, this question arose as I was trying to solve another question in the book. My intuition tells me the answer is no, but I haven't been able to come up with a satisfying proof. Any help would be appreciated.

Edit: in particular, I mean for $k'$ to be a proper subfield of $k$, and the quotient of $A$ to be isomorphic to $k'$ as a ring.

Answer 1

Sure, this is possible. For instance, let $L$ be any field and let $k=L(x_0,x_1,\dots)$ be a field of rational functions in infinitely many variables and $k'=L(x_1,x_2,\dots)$. Then $k'$ is isomorphic to $k$ (just shift the variables), so we can take $A=k$ and the quotient by the zero ideal.

Answer 2

The answer is no. Let $k \to A$ be the structure map and let $I$ be an ideal of $A$. Then the composition $k \to A \to A/I$ is nonzero and therefore injective.

EDIT: As Eric Wofsey points out this is in fact possible, as long as $k'$ contains a copy of $k$, which shouldn't be possible in many cases we care about (e.g if $k=\mathbb{C}$ and $k'$ is the real numbers)