Change of Variables -- is it always an isomorphism?

Question

I think I'm fundamentally confused about what it means to make a change of variables (in the context of an affine/projective variety).

From the usage I see, it seems to be some sort of transformation $\varphi : k[x_1,..,x_n] \rightarrow k[x_1,..,x_n]$ that relabels a polynomial expression as a variable, e.g. $\varphi(y^2/2 + x) = X$.

Do we also require a change of variables to be an isomorphism of $k[x_1,..,x_n]$? (Does anyone have a reference for the definition?)

For a concrete example, I realized my confusion reading the answers here: Hartshorne exercise 1.1 (c). I'm not sure I understand why the change of variables used there in the second answer leads to an isomorphism.

Answer 1

The concept of "changing variables" is a blanket name for the general technique of replacing some variables with other (more convenient) expressions. I'm not sure I've ever seen a rigorous definition in a text - most of the time anyone who's using it will explain enough to clue in the readers as to what they're doing.

Let's look at your concrete troubles. In both answers in your linked post, the specific changes of variable are of the form $$ x\mapsto ax+by+p$$ $$y\mapsto cx+dy+q$$ for $a,b,c,d,p,q\in k$ and each author assumes various facts which imply $ad-bc\neq 0$ in each transformation they discuss. These transformations are all automorphisms of $k[x,y]$, and we can determine their explicit inverse: all such transformations are in fact "affine transformations" of the plane $k^2$, and can be recorded in a matrix as $$\begin{pmatrix} a & b & p \\ c & d & q \\ 0 & 0 & 1 \end{pmatrix},$$ whence composition of transformations corresponds to multiplication of matrices, and inversion corresponds to inverting the matrix representing the transformation. The explicit inverse therefore corresponds to the transformation represented by the matrix $$\begin{pmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc} & \frac{bq-dp}{ad-bc} \\ \frac{-c}{ad-bd} & \frac{a}{ad-bc} & \frac{cp-aq}{ad-bd} \\ 0 & 0 & 1 \end{pmatrix}. $$

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Back to the general picture, I have seen "change of variables" used for various birational equivalences relating to elliptic curves which are not necessarily isomorphisms. I would say that generally a "change of coordinates" on an affine variety means picking different generators for the coordinate algebra - this will necessarily be an isomorphism. A "projective coordinate change" usually means acting by some automorphism of $\Bbb P^n$, which are all linear maps represented by some element in $PGL(n+1)$. So the phrasing "change of coordinates" should always imply that it's an isomorphism, while "change of variables" may not in some cases.

Answer 2

A change of variables always means an isomorphism. In the case of the answer to the linked question, we have $$p(x,y) = x^2 + a(y)x+b(y) = \left(x+ \frac{a(y)}{2}\right)^2 + b(y)-\frac{a(y)^2}{4}= \left(x+ \frac{a(y)}{2}\right)^2 + B(y) $$ where we put $B(y) = b(y)-\frac{a(y)^2}{4}$. The change of variables is just $$(z,w) = \varphi(x,y) = \left(x+ \frac{a(y)}{2}, y \right)$$ which is an isomorphism $k[x,y] \rightarrow k[z,w]$ if the characteristic of $k$ is different from two.

In these new variables the curve is given by $( p \circ \varphi^{-1}) (z,w) = z^2 + B(w) $ as it is claimed.