Exercise in Hartshorne

Question

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

Answer 1

Here's another approach. Let's reconsider the first exercise.

Let $\phi:k[x,y] \rightarrow k[t]$ be the the map given by $g(x,y) \mapsto g(t,t^2).$ Then $\phi$ is an epimorphism which contains in it's kernel the polynomial $y -x^2.$ We wish to show $\ker \phi = (y - x^2).$

First note as the image of $\phi$ is an integral domain, the kernal of $\phi$ is prime. Furthermore, the $\dim k[t] = 1$ and $\dim k[x,y] =2.$ Therefore, it must be the case that $ht(\ker \phi) = 1.$ We show $(y -x^2)$ is also a prime of height $1.$

Observe that the polynomial $x^2 - y \in k[y][x]$ satisfies eisenstein's criterion at the prime $(y).$ Hence, $x^2 - y$ is irreducible in $k[y,x].$ As $k[y,x]$ is a UFD, it follows $x^2 - y$ is prime.

Hence, $(x^2 - y)$ is a prime of height $1.$ Given the containment $(y^2 - x) \subset \ker \phi,$ it must be the case $(y^2 -x) = \ker \phi.$

For the second exercise, one can use this same argument to show $k[x,y,z]/(x^3 - z) \cong k[x,y].$ And so, $k[x,y,z]/(x^3 - z, x^2 - y) \cong k[x,y]/(x^2 - y) \cong k[x].$

Answer 2

Following Gooz and Dylon Moreland's comment I came up with a proof, that is not as elegant as jspecter's but more elementary.

I use the following fact from Lang chapter 4 theorem 1.1.: Let $f(x),g(x) \in R[x]$ and suppose that the leading term of $g(x)$ is invertible in $R$. Then $f(x)=g(x)d(x)+r(x)$ where the degree of $r(x)$ is smaller than that of $g(x)$.

I will directly consider exercise 2 and let $f(x,y,z)$ be a polynomial the kernel of $\phi$, where $\phi$ is the homomorphism $k[x,y,z]\to k[t]$ given by $x \mapsto t$, $y \mapsto t^2$ and $z\mapsto t^3$. We wish to show that $(x^2-y,x^3-z)$ is the kernel.

Consider $f(x,y,z)$ as a polynomial in $k[x,y][z]$. Then the leading term in $x^3-z$ (in $z$) is invertible in $k[x,y]$, so we write $f(x,y,z) = (x^3-z)d(x,y,z)+r(x,y,z)$. Then either $r(x,y,z)=0$ and we are done, or it has degree less than $(x^3-z)$ has, so $r(x,y,z)$ is in fact a polynomial only in $y$ and $x$. We again apply the result from above and write $r(x,y)=(x^2-y)d'(x,y)+r'(x,y)$. And we are done if $r'(x,y)=0$. Otherwise with the same reason as before $r'$ is polynomial only in $x$.

So now we have $$ f(x,y,z) = (x^3-z)d(x,y,z) + (x^2-y)d'(x,y) +r'(x).$$ If we now apply $\phi$ we see that $0=r'(t)$ and this is only possible if $r'=0$, so we have $$ f(x,y,z) = (x^3-z)d(x,y,z) + (x^2-y)d'(x,y) \in (x^2-y,x^3-z). $$

Answer 3

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.

Answer 4

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.