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So in Hartshorne I exercise 1.11 we have the set $Y = \{(t^3, t^4, t^5) : t\in k\} \subset \mathbb{A}^3$ ($k$ algebraically closed) and we have to show that $\mathfrak{p} = I(Y)$ is a prime of height $2$ that cannot be generated by two elements. There's a few questions here about it, see here, here and here for instance. In this case one shows that $\mathfrak{p} = (y^2 - xz, x^3 - yz, z^2 - x^2 y)$.

I worked out the exercise, but now I have another question: is $Y$ the intersection of two surfaces? That is, are there irreducible polynomials $f, g \in k[x,y,z]$ such that $Y = V(f, g)$ - or concretely, such that $\mathfrak{p} = \sqrt{(f,g)}$? Do we need them to be irreducible?

What I had thought so far is: if we localize $A = k[x,y,z]$ at $\mathfrak{p}$, we get a Noetherian local ring $A_\mathfrak{p}$ of dimension $\dim A_\mathfrak{p} = \operatorname{ht} \mathfrak{p} = 2$ and with maximal ideal $\mathfrak{p}^e$. So by e.g. Atiyah-MacDonald 11.14 there is a $\mathfrak{p}^e$-primary ideal $\mathfrak{a} \subset \mathfrak{p}^e \subset A_\mathfrak{p}$ generated by two elements. I think this means the corresponding ideal $\mathfrak{a}^c \subset A$ is $\mathfrak{p}$-primary, but can we translate those 2 generators of $\mathfrak{a}$ into 2 generators of $\mathfrak{a}^c$? Or perhaps I should instead be looking at generators of $\mathfrak{p}^e/(\mathfrak{p}^e)^2$ over $k$ and using some form of Nakayama's lemma? But I'm not sure if $A_\mathfrak{p}$ is regular local.

On the other hand, I don't know how one would compute explicit pairs $f,g$. Is this where one needs to bring in so-called Gröbner basis techniques or elimination theory?

I haven't made it super far through Chapter I yet, so maybe I'm missing some standard result for set-theoretic complete intersection curves in $\mathbb{A}^3$ or $\mathbb{P}^2$. I recall that it is a famous open problem in $\mathbb{P}^3$.

csaltachin
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  • You're right, that's a typo. Thanks for catching that. – csaltachin Jun 07 '25 at 10:47
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    Yes, this curve is a set- theoretic complete intersection. If you can't find it in literature, I will explain when I have more time. – Mohan Jun 07 '25 at 17:37
  • This question has already been discussed on this site, search first, ask after. – Sasha Jun 07 '25 at 18:09
  • @Mohan Thank you, I'll keep looking around but I'd appreciate any pointers. – csaltachin Jun 07 '25 at 18:53
  • @Sasha Note that what I ask is different from what the exercise asks. I've looked at a few questions in math.se about the exercise in the last couple days -- including those I linked -- but I couldn't find anyone writing $Y$ as the intersection of two surfaces, or listing 2 explicit generators for a $\mathfrak{p}$-primary ideal. – csaltachin Jun 07 '25 at 19:00

1 Answers1

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I want to apply the algorithm from Bresinsky 1979 (jstor), but first I set the stage by projecting the $5$-uple embedding $\nu_5$ of ${\Bbb P}^1$ to ${\Bbb P}^3$ recovering the situation of the exercise by looking in the affines $s=w=1,$ in Macaulay2:

S=QQ[s,t]
R=QQ[w,x,y,z]
nu5p=map(S,R,{s^5,s^2*t^3,s*t^4,t^5})
dh=map(QQ[x,y,z],R,{1,x,y,z})
dh ker nu5p -- ideal(y^2-x*z,x^2*y-z^2,x^3-y*z) -- f2, f1, f3

I apply the algorithm with $x_1=z,x_2=y,x_3=x$, checking first that we also get the line $x_1=x_2=0$ when using only $f_1,f_2$ as generators

primaryDecomposition ideal(y^2-x*z,x^2*y-z^2) -- {ideal(z,y), ideal(y^2-x*z,x^2*y-z^2,x^3-y*z)}

Finding the $g$ of the algorithm amounts to adding or subtracting multiples of $f_1$ from $f_2^2$ and dividing by $y$

factor((y^2-x*z)^2+x^2*(x^2*y-z^2)) -- y*(x^4+y^3-2*x*y*z)

Note that the plane $y=0$ contains the line $y=z=0.$ Then the ideal we seek is $I=\langle f_1,g \rangle$

I=ideal(x^2*y-z^2, x^4+y^3-2*x*y*z) -- f1, g
(y^2-x*z)^2%I -- 0
(x^3-y*z)^2%I -- 0

Plotting this there seems to be a tangency at the intersection, indicating the multiple structure on the curve

g,f_1

Jan-Magnus Økland
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