I am trying to solve the following exercise:

Find the Zariski-closure of the set $\{(a^2,a^3,a^4)|a \in \mathbb{C}\} \subset \mathbb{C}^3$.

Let's call the set above $A$.

Now, my original idea was to look at the morphism $\varphi:\mathbb{C} \to \mathbb{C}^3$ defined explicitly by $a \mapsto (a^2,a^3,a^4)$ which induces a $\mathbb{C}$-algebra homomorphism $\phi:\mathbb{C}[x,y,z] \to \mathbb{C}[u]$ explicitly defined by $x \mapsto u^2,y \mapsto u^3,z \mapsto u^4$. We note that $\text{Im}(\varphi) = \{(a^2,a^3,a^4) | a \in \mathbb{C}\} = A$, and by theorem 15.2.16 in Dummit & Foote (3rd edition) we know that the kernel of $\phi$ is $I(\text{Im}(\varphi))$ and hence that the zero-set of the kernel of $\phi$ is the zariski-closure of $A$.

Now by theorem 15.1.8 in Dummit & Foote, adapted to the current setting, we have the following proposition: Let $R = \mathbb{C}[x,y,z,u]$ with the monomial ordering $u > x > y > z$ and let $\mathcal{A}$ be the ideal generated by $x-u^2,y-u^3,z-u^4$. Let $G$ be the reduced gröbner basis for $\mathcal{A}$ with respect to the given monomial ordering. Then we find that $\text{ker} \ \phi = \mathcal{A} \cap \mathbb{C}[x,y,z]$ and that the elements of $G$ in $\mathbb{C}[x,y,z]$ generate $\text{ker} \ \phi$.

Now, if we set $g_1 = x-u^2,g_2 = y-u^3,g_3 = z-u^4$ we have $\mathcal{A} = (g_1,g_2,g_3)$.

Applying the buchberger algorithm, we immedietaly find that $$S(g_1,g_2) = y-xu \equiv 0 \ \text{mod} \ \ \mathcal{A}, S(g_1,g_3) = z -xu^2 \equiv 0 \ \text{mod} \ \mathcal{A},S(g_2,g_3) = z-yu \equiv 0 \ \text{mod} \ \mathcal{A}.$$

Now, here I am stuck and don´t know what to do. What I want to do is to find a reduced gröbner-basis, but according to buchbergers algorithm I am done. How does one proceed to find the reduced gröbner-basis; is there a general algorithm? Or is there another way to solve this that does not use a gröbner-basis?

(I´d be more happy to receive a hint than a full solution).

A similar computation is done here, but I don´t know exactly how they arrive at $x^3-y^2$ by the way of computing the reduced gröbner-basis, since this is not spelled out in detail:

enter image description here

Edit: More detailed calculation of the buchberger algorithm:

Fixing the monomial order $u > x > y > z$ we have that $\text{LT}(g_1) = -u^2$, $\text{LT}(g_2) = -u^3$ and $\text{LT}(g_3) = -u^4$. We then have that $\text{LCM}(\text{LT}(g_1),\text{LT}(g_2)) = u^3, \text{LCM}(\text{LT}(g_1),\text{LT}(g_3)) = u^2$ and $\text{LCM}(\text{LT}(g_2),\text{LT}(g_3)) = u^4$. Then one finds that $$S(g_1,g_2) = \frac{u^3}{-u^2}(x-u^2)-\frac{u^4}{-u^3}(y-u^3) = -u(x-u^2)+(y-u^3) = y-ux$$ and similar computations for $S(g_1,g_3)$ and $S(g_2,g_3)$ gives us the result.

$\underline{Note}$: the $\text{LCM}$ here is the monic $\text{LCM}$.

  • 2
    $\begingroup$ You know that xz-y2 and z-x2 are in the kernel, and maybe look at dimensions $\endgroup$
    – usr0192
    Commented Jan 7 at 4:58
  • $\begingroup$ @usr0192 You might want to Mathjax to type your comment. $\endgroup$
    – KaleBhodre
    Commented Jan 7 at 5:13
  • $\begingroup$ May be this related thread helps you? $\endgroup$ Commented Jan 7 at 5:14
  • $\begingroup$ I answered this question here: math.stackexchange.com/questions/4833927/… $\endgroup$
    – Kenta S
    Commented Jan 7 at 6:24
  • $\begingroup$ Thanks @KentaS. How did you arrive at $Z(x^2-z,x^3-y^2)$ in your solution? That is, what led you towards that, it is not spelled out, and so is not that interesting unless you can explain how you reasoned to get $Z(x^2-z,x^3-y^2)$. $\endgroup$
    – Ben123
    Commented Jan 7 at 6:37

1 Answer 1


Buchberger's alogrithm doesn't say you're done quite yet! I think a point of confusion might have been that in Buchberger's algorithm, we aren't analyzing whether $$S(g_1,g_2) \equiv 0 \mod \mathcal A,$$ since $S(g_1,g_2)$ is definitely in the ideal by definition (it's generated by $g_1,g_2$). Instead, we will to compute $$S(g_1,g_2) \equiv ux-y \mod G,$$ which is the notation Dummit and Foote uses for the remainder of a polynomial after general polynomial division (page 320).

Following your notation, denote $$g_1 = x - u^2, \qquad g_2 = y-u^3, \qquad g_3 = z-u^4,$$ and for now define $G = \{g_1, g_2, g_3\}$ the ordered set of polynomials. As per the algorithm, we compute $$S(g_1,g_2) = ux-y,$$ and let's denote it $g_4$. Buchberger says to try dividing it by $G$, and if we get something nonzero, we add it to our list of generators.

We divide by $G$. Starting with $LT(g_4) = ux$, we go through our existing generators $g_1,g_2,g_3$ and see if their leading terms divide it. They do not, so we at least have the remainder $ux$. Removing the leading term from $g_4$, we see if any of our generators $g_1,g_2,g_3$ divide $LT(g_4-ux)=LT(-y)=-y$. They do not, so we have remainder $ux-y$ after dividing by $G$.

So, we add in $g_4$ to our set, giving us $G=\{g_1,g_2,g_3,g_4\}$. Now, we repeat this process, trying $S(g_1,g_3)$, and so on.

You can do this exercise by continuing this process to get a Grobner basis, and then reducing the Grobner basis by continually replacing a generator with its remainder after general polynomial long division by the other generators, as described before Theorem 27 in Section 9.6 of Dummit and Foote. I'll include a little bit more of the computation.

So far, we have $$G = \{x-u^2, y-u^3, z-u^4, ux-y\}.$$ Next, we compute $S(g_1,g_3) = u^2 x - z$. Dividing this by $G$, we get $uy-z$, so our new generators are $$G = \{x-u^2, y-u^3, z-u^4, ux-y, uy-z\}.$$ Next, $S(g_2,g_3)=uy-z$, so this has remainder $0$ after dividing by $G$.

Next, $S(g_1,g_4)= -uy + x^2$, and dividing by $G$ (in particular $uy-z$), we get $x^2-z$, so our new generators are $$G = \{x-u^2, y-u^3, z-u^4, ux-y, uy-z, x^2-z\}.$$ Next, $S(g_1,g_5)=xy-uz$, and dividing by $G$, we get again $xy-uz$. Thus, $$G = \{x-u^2, y-u^3, z-u^4, ux-y, uy-z, x^2-z, xy-uz\}.$$

We will continue until all pairwise $S(g_i, g_j) \equiv 0 \mod G$. Then, we will reduce the Grobner basis, and apply Dummit and Foote Theorem 15.1.8 .

We could finish the computation using Macaulay2

R = QQ[u,x,y,z, MonomialOrder => Lex];
I = ideal(x-u^2, y-u^3, z-u^4);
gens gb I;

which gives $$\mathcal A = (y^4-z^3, xz-y^2, xy^2-z^2, x^2-z, uz-xy, uy-z, ux-y, u2-x),$$ so intersecting $\mathcal A \cap \mathbb C[x,y,z]$, we get $$\ker\phi = (y^4-z^3, xz-y^2, xy^2-z^2, x^2-z)$$

As some of the comments pointed out, you can also approach this problem by "guessing" elements that cut out the image of $\varphi$. For $x=t^2$ and $y=t^3$, one relation that is true is that $x^3 = t^6 = y^2$, so $x^3-y^2 \in I(\{(a^2,a^3,a^4) \in \mathbb C^3\})$. One can keep guessing "true facts" about the Zariski closure, and then to show that your relations are sufficient you'll often have to do something with dimension or something about some sort of ad-hoc (as opposed to Grobner basis stuff) polynomial long division. You might want to check out this answer.

  • $\begingroup$ Thanks for a thorough answer. I note that your answer is different from Kenta:s answer in the link in the comments, unless it is the case that they generate the same ideal (I have not checked). I am quite curious how many computations took place in that last step you did, since that tells me the feasibility of using this method by hand when you have examples of roughly this size (say on an exam). Also, you are of course correct that $S(g_i,g_j) \equiv 0 \ \text{Mod} \ \mathcal{A}$ by definition. $\endgroup$
    – Ben123
    Commented Jan 7 at 7:28
  • $\begingroup$ Both generate the same ideal, so we do have the same answer. My ideal contains Kenta's, since it has $x^2-z$ and $y^2-x^3$ can be gotten from $xz-y^2$, replacing $z$ with $x^2$ using $x^2-z$. Similarly Kenta's ideal contains mine (show that all my generators are 0 modulo their ideal). I don't believe there are that many calculations left---I got tired and use Macaulay2 haha---but Kenta's solution does seem more time efficient for an exam. There are more manageable small examples that could feasibly be on an exam, maybe Dummit and Foote page 667 example 1. $\endgroup$
    – Joseph
    Commented Jan 7 at 7:43

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