TITLE: Exercise in Hartshorne
QUESTION [10 upvotes]: I started reading Hartshorne. Already in the first exercises I stumble across problems.
Basically excerise 1.1 ask to prove that $k[x,y]/(y-x^2)$ is isomorphic to a polynomial ring in one variable. Well ok, so I tried the following, define $k[x,y]\to k[t]$ by $x \mapsto t$
and $y \mapsto t^2$. This is obviously a homomorphism and also $y-x^2$ is in the kernel. My problem was to show that the kernel is nothing more than $(y-x^2)$ though this seems kind of clear but I had some trouble proving it rigorously.
I did the following Let $\sum_{i,j} a_{i,j}x^iy^j$ be in the kernel. Then the image is
$\sum_{i,j} a_{i,j}t^it^{2j}=0$. First note that $\sum_{i,j} a_{i,j}x^ix^{2j}$ is also in the kernel, since it has the same image as $\sum_{i,j} a_{i,j}x^iy^j$. But we also see that
$\sum_{i,j} a_{i,j}x^ix^{2j}$ is actually zero, since it is the same as $\sum_{i,j} a_{i,j}t^it^{2j}=0$ only with variables renamed. Thus
$$\sum_{i,j} a_{i,j}x^iy^j=\sum_{i,j} a_{i,j}x^iy^j-\sum_{i,j} a_{i,j}x^ix^{2j}=
\sum_{i,j} a_{i,j}x^i(y^j-x^{2j})$$, and it is well known that $a-b$ divides $a^j-b^j$.
Thus the sum is divisble by $y-x^2$ and we are done. Pretty complicated proof for something
that seems almost obvious.
So my problem comes when doing exercise 1.2
I have to show that $k[x,y,z]/(x^2-y,x^3-z)$ is isomorphic to a polynomial ring in one variable.
So again I define a homomorphism $k[x,y,z]\to k[t]$ by $x \mapsto t$, $y \mapsto t^2$ and
$z\mapsto t^3$. The ideal $(x^2-y,x^3-z)$ is obviously contained in the kernel. But how do I show that it is all of that. Taking the same approach as above seems to lead to an even more complicated proof of an "obvious fact". 
So how can I prove 1.1 in a nicer technique that also applies to 1.2

REPLY [6 votes]: Let me give a completely elementary argument.
Let $A=k[X,Y]/(X^2-Y)$, and let $x$ and $y$ be the images of $X$ and $Y$ in $A$. There is a algebra map $k[X,Y]\to k[t]$ such that $X\mapsto t$ and $Y\mapsto t^2$; the element $X^2-Y$ is in its kernel, so the map descend to an algebra map $\alpha:A\to k[t]$ such that $\alpha(x)=t$ and $\alpha(y)=t^2$. Now obviously there is a map $\beta:k[t]\to A$ such that $\beta(t)=x$.
The composition $\alpha\circ\beta$ is the identity of $k[t]$: since it is an algebra map, it is enough to compute its action on the generator $t$, and $\alpha(\beta(t))=\alpha(x)=t$. Similarly, $\beta\circ\alpha$ is the identity in $A$: $\beta(\alpha(x))=\beta(t)=x$ and $\beta(\alpha(y))=\beta(t^2)=x^2=y$. 
We conclude that $\alpha$ and $\beta$ are isomorphisms.
The exact same argument deals with your other example.

REPLY [5 votes]: At the risk of stating the obvious, rather than computing the kernel of your homomorphism, you could just write down its inverse: $t \mapsto x$. The argument then boils down to showing that
$$f(x,y,z) \equiv f(x, x^2, x^3) \mod{\left\langle y - x^2, z - x^3 \right\rangle}$$
which is true because equivalence modulo an ideal is a congruence relation. e.g. so we can use $y \equiv x^2$ to make a substitution.