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

  • 2
    $\begingroup$ you should use division with remainder, I think. So in the first exercise you have a polynomial $f(x,y)$ such that $f(t,t^2) =0$. Division with remainder should help you see that $f(x,y)$ contains a factor $(y-x^2)$. $\endgroup$
    – Gooz
    Oct 1, 2011 at 13:43
  • 1
    $\begingroup$ On a less elementary tune: your map is obviously onto, so the kernel must be a prime ideal. The ideal $(y-x^2)$ is prime since the generator is irreducible. If the kernel were bigger, it would be a prime ideal strictly containing $(y-x^2)$, thus ... $\endgroup$ Oct 1, 2011 at 14:08
  • $\begingroup$ @Gooz: But $k[x,y]$ ist not euclidean. How do you do division with remainder then? $\endgroup$
    – wood
    Oct 1, 2011 at 14:12
  • $\begingroup$ @AndreaMori. In fact, I don't get it. It is possible that a prime ideal strictly contains another one. Imagine the second example. the map is onto and $(y-x^2)$ is prime, but still there is another prime ideal $(y-x^2,z-x^3)$ strictly containing it. $\endgroup$
    – wood
    Oct 1, 2011 at 14:20
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    $\begingroup$ @claudi85 I think Prof Mori might have some argument about the dimension in mind. In any event, while this is often not mentioned, Euclidean division of polynomials can be done in any commutative $R[X]$ as long as you divide by a polynomial whose leading coefficient is a unit in $R$. There is a proof in Lang's Algebra, but if you know the proof when $R$ is a field then that should carry over. $\endgroup$ Oct 1, 2011 at 14:26

4 Answers 4


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, $k[x,y,z]/(x^3 - z, x^2 - y) \cong k[x,y]/(x^2 - y) \cong k[x].$

  • $\begingroup$ uses some machinery but a very elegant proof. thanks $\endgroup$
    – wood
    Oct 1, 2011 at 15:29

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 $$ f(x,y,z) = (x^3-z)d(x,y,z) + (x^2-y)d'(x,y) \in (x^2-y,x^3-z). $$


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.


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.


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