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I am working on the following problem: Let $Y \subset A^3$ be the set $Y={(t,t^2,t^3)|t\in k}$
($A^3$ is the affine 3-space over $k$ an algebraically closed field.) Show that $Y$ is an affine variety of dimension 1. Find generators for the ideal $I(Y)$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$.

I have found an answer which goes like this:

Clearly, $A(Y)=k[x,y,z]/(z-x^3, y-x^2)$. Note that $A(Y) = k[x,x^2,x^3]= k[x]$ so Y is irreducible, and $Y$ is an affine variety. and dim $Y$ = dim $A(Y)$ = dim $k[x]$ = 1 and $I(Y) = (z-x^3,y-x^2)$

I am finding it difficult to understand this answer, it doesn't even look like it has addressed each part of the answer. In particular, how does one show $A(Y)=k[x,y,z]/(z-x^3, y-x^2)$?

I would also greatly appreciate a better way to approach this.

Mathemphetamine
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1 Answers1

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In order to show that $A(Y) = k[x, y, z]/(z-x^3, y-x^2)$, the simplest way is to show that $Y = Z(z-x^3, y-x^2)$, i.e. the variety in $\Bbb A^3_k$ that has the coordinate ring $ k[x, y, z]/(z-x^3, y-x^2)$ has the same underlying set of points as the image of the map $t \mapsto(t, t^2, t^3)$.

This is done by showing inclusion both ways. First of all, it's easy to check that $Y \subseteq Z(z-x^3, y-x^2)$, since for any $t \in k$ we have $t^3-t^3 = 0$ and $t^2 - t^2 = 0$.

For the other inclusion, say we have a point $(a, b, c) \in Z(z-x^3, y-x^2)$. That means that $c - a^3= 0$ and $b - a^2 = 0$, or in other words, $b = a^2$ and $c = a^3$. Then clearly $a \mapsto (a, b, c)$ by the parametrisation that defines $Y$, so $(a, b, c) \in Y$.

As for how to guess at $(z-x^3, y-x^2)$ as the defining ideal in the first place, it's mostly practice. There are, however, a few pointers to get you started: Since $Y$ is a curve in $\Bbb A^3_k$, you should try to look for two generators as a start. Some times you might need more (like for instance $t \mapsto (t^3, t^4, t^5)$, which corresponds to the ideal $(x^5 - z^3, x^4 - y^3, y^5 - z^4)$). Other than that, you just stare at the parametrisation and write down any relations that you see fit, in as low degree as possible. Some times you write too many, some times the presence of one generator means you can simplify another.

Arthur
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  • Is $(z-x^3, y-x^2) = (z-x^3, y-x^2, y^3- z^2)$ ? The inclusion "$\subseteq$" is obvious, but I cannot see immediately if $y^3 - z^2 \in (z-x^3, y-x^2)$ or not....

    The reason I ask is because I naively wrote down $(z-x^3, y-x^2, y^3- z^2)$ as the defining ideal of the twisted cubic curve, the quotient by this ideal is still $\cong k[t]$.

     – cat Jul 26 '16 at 05:11
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    @cat Note that $$y^3 - z^2 + (y^2 + x^2y + x^4)(x^2 - y) - (z +x^3)(x^3-z) = 0$$Working it out directly took a small amount of work, but if you divide out by $(x^2-y,x^3-z)$ first, you can see directly that $y^3-z^2 \simeq x^6-x^6 = 0$. – Arthur Jul 26 '16 at 05:35
  • I see... So in general, can we say that: If $I\subset J$ are ideals of a (let's just say) commutative ring $R$, then $R/I \cong R/J$ implies $I =J$ ? – cat Jul 26 '16 at 05:43
  • @cat Yes. For a fixed ideal $I$ there is a one-to-one correspondence between ideals $J \supseteq I$ and ideals $J' \subseteq R/I$. This is often one of the first theorems proven after introducing quotient rings in any textbook. – Arthur Jul 26 '16 at 05:45
  • Why did you choose the polynomials $z-x^3$ and $y-x^2$? Why not $xy-z$ and $xz-y^2$? – user5826 Mar 31 '21 at 22:19
  • @user46372819 It was almost five years ago. I honestly can't tell you why I made one choice over another back then. I don't think anyone could. But looking back at it, $z-x^3$ and $y-x^2$ look a lot prettier to me. They have more separation between the variables and seem to give quotients that are easier to work with. Also, it might be just because it's late and I'm tired, but it's not immediately obvious to me that $(z-x^3, y-x^2)\subseteq (xy-z, xz - y^2)$. – Arthur Mar 31 '21 at 22:42
  • (OK, after actually picking up a pen and a piece of paper and scribbling for a minute, I see now that the two ideals probably are equal. I don't trust myself this late, though.) – Arthur Mar 31 '21 at 22:48