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Show the set of points $X = \{(t^3, t^4, t^5) \}$ with $t\in k$ is closed in $\mathbb A^{3}$ and find three generators of $\mathcal{I}(X)$.

This is a homework question, so please don't provide full answers. I've just started algebraic geometry and am not familiar with techniques to use at all.

I have that the canonical map $k[x,y,z] \rightarrow k[t^3, t^4, t^5]$ has kernel $\mathcal{I}(X)$. Since it's surjective then $k[x,y,z]/ \mathcal{I}(X) \simeq k[t^3, t^4, t^5]$. But $k[t^3, t^4, t^5]$ is an integral domain so $\mathcal{I}(X)$ must be prime. Then $X$ is irreducible (by the correspondence $\mathcal{I}(-)$ between prime ideals and irreducible subsets).

But I have no idea how to find the generators.

Looking around it seems that the generators are $xz-y^2, yz-x^3, z^2-x^2y$, but I don't know how these are deduced or to prove that it is indeed $\mathcal{I}(X)$ that they span (of course it is contained in $\mathcal{I}(X)$).

Any help, please?

Thanks in advance.

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  • $\begingroup$ Do you know anything about Gröbner bases? I think finding the generators for this ideal is a standard problem there... $\endgroup$ Feb 10, 2014 at 0:35
  • $\begingroup$ No, I don't know about Gröbner bases, unfortunately. $\endgroup$
    – user68193
    Feb 10, 2014 at 22:27

3 Answers 3

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Edit: As pointed out in the comments, my original answer was incorrect. I have now updated it to something that I believe works.

What we want to do is find some ideal $I$ such that $Z(I) = X$, to show that it is closed.

Something that is helpful to notice is that if $(x,y,z)\in X$ isn't $0$, then $\frac{z}{y}=t=\frac{y}{x}$. So any element of $X$ satisfies the equations $ZX=Y^2$ and $ZX^5=Y^5$ i.e. lies in the affine variety generated by the two polynomials $ZX-Y^2$ and $ZX^5-Y^5$.

To show that this variety is exactly $X$, you need to show that for a given point satisfying both of the given equations, you can find some $t$ such that the point is of the form $(t^3,t^4,t^5)$ (hint, look at how we derived the two equations from $t$ and try and reverse the procedure.)

In general, showing that $I(X)$ is generated by a given set of generators is hard, but provided you have a small number of generators where some variables occur to low degrees, there is a method that will sometimes work, it is essentially a glorified division algorithm:

We try to show that $I(X)=\langle Z^2-X^2Y, Y^2-XZ, X^3-YZ\rangle$*. Clearly each of the three given elements lies in $I(X)$, so we just need to show that any element of $I(X)$ is lies in the right hand ideal. First suppose $f(X,Y,Z)$ is in $I(X)$. Then we can look at the monomials occurring in $f$, and look in particular at how many times $Z$ appears in each monomial. In each monomial, $Z$ might not appear at all, $Z$ might appear once, or $Z$ might appear two or more times. This means that we can split $f$ up into simpler polynomials:

$$f(X,Y,Z)=Z^2f_2(X,Y,Z)+Zf_1(X,Y)+f_0(X,Y)$$

where $f_0$ is the sum of monomials in $f$ where $Z$ doesn't appear, $Zf_1$ is the sum of monomials in $f$ where $Z$ appears once, and $Z^2f_2$ is the sum of monomials in $f$ where $Z$ appears at least twice. You can also think of this as applying the division algorithm to $f$ when you consider $f$ as a polynomial in $Z$ with coefficients in $k[X,Y]$, and you are dividing $f$ by $Z^2$ to get remainder $Zf_1(X,Y)+f_2(X,Y)$.

Now we can re-write the above equation as follows:

$$f(X,Y,Z)=(Z^2-X^2Y)f_2(X,Y,Z)+X^2Yf_2(X,Y)+Zf_1(X,Y)+f_0(X,Y).$$

But since $I(X)$ is an ideal and $f(X,Y,Z), (Z^2-X^2Y) \in I(X)$, this means that:

$$f(X,Y,Z) - (Z^2-X^2Y)f_2(X,Y) = Zf_1(X,Y)+f_3(X,Y) \in I(X)$$

where $f_3(X,Y) = X^2Yf_2(X,Y) +f_0(X,Y)$, a polynomial only involving powers of $X$ and $Y$.

Now we can repeat the process with $f_1(X,Y)$ and $f_3(X,Y)$, but instead of looking at how many times $Z$ appears in the monomial of each of these, we now look at how many times $Y$ appears. Then we can write $f_1(X,Y)$ as a sum of three polynomials in just the single variable $X$, with one of them multiplied by $Y$ and one of the multiplied by $Y^2$. We can then do a similar thing for $f_3$. Using these expressions for $f_1$ and $f_3$, and a similar argument to what we used to show $Zf_1(X,Y)+f_3(X,Y) \in I(X)$, we can show that some polynomial of the form:

$$XZ^2g_1(X)+ZYg_2(X)+Zg_3(X)+XZg_4(X)+Yg_5(X)+g_6(X)$$

lies in $I(X)$. But by writing $XZ^2g_1 = (Z^2-X^2Y)Xg_1 + X^3Yg_1$ and by realizing that e.g. $Xg_4(X)$ is still a polynomial in $X$, we can deduce that a polynomial of the form:

$$h(X,Y,Z) = YZh_1(X) + Zh_2(X)+Yh_3(X)+h_4(X)$$

lies in $I(X)$. The natural continuation would be to repeat the process with each of the $h_i(X)$ and the third generator $X^3-YZ$, and then multiply out and deduce that some polynomial $f'(X,Y,Z)$ which is a sum of monomials obtained by multiplying one element of the set $\{YZ,Z,Y,1\}$ with one element of the set $\{YZ,X^2,X,1 \}$ lies in $I(X)$. But this would mean that $f'$ vanishes on $X$, and so for all $t$, $f'(t^3,t^4,t^5) = 0$. Assuming we have an infinite (and in particular algebraically closed) field, this then implies that the polynomial in $t$, $f'(t^3,t^4,t^5)$ is then the zero polynomial. But then you can check by hand that all of the monomials that appear in $f'$ give different powers of $t$ when we substitute $X = t^3, Y = t^4, Z = t^5$, so that $f'$ must have been the zero polynomial also.

Thinking about what we have done here, you should be able to see that $f'$ is in some sense a remainder of $f$ when divided by the three generators. Explicitly, what we did was write $$f(X,Y,Z) = (Z^2-X^2Y)\bar{f_1}(X,Y,Z)+(Y^2-XZ)\bar{f_2}(X,Y,Z)+(X^3-YZ)\bar{f_3}(X,Y,Z)+f'(X,Y,Z)$$ and then deduced that $f'(X,Y,Z) = 0$. In other words we have shown that any element of $I(X)$ lies within $(Z^2-X^2Y,Y^2-XZ,X^3-YZ)$, so this ideal is all of $I(X)$.

Remark: To make things a bit simpler computationally, we could have substituted powers of $t$ for $X,Y$ & $Z$ into our expression for $h$, and looked at the possible powers of $t$ appearing in each of the four simpler polynomials in our expression of $h$ modulo $3$. Since $Zh_2(X)$ is the only part contributing terms of order equal to $2\pmod 3$ and $Yh_3(X)$ is the only part contributing terms of order equal to $1\pmod 3$, we must have that $Zh_2(X)=0=Yh_3(X)$. We may now decompose $h_1(X)$ and $h_3(X)$ as before, but our remainder $f'$ now consists of fewer monomials, so it is easier to check that $f'$ is $0$.

*The real question is, why did we choose these generators? If you use the method described it should be reasonably clear why we want something similar to the given ones. What we look for in possible choices of generators are elements of $I(X)$ which have a small number of terms (here each generator has two terms, but if one term were possible that would be better) and where one of the terms is a power of $X,Y$ or $Z$, and preferably as low a power as possible.

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  • $\begingroup$ I've understood everything, but I end up with a polynomial $Yg(X)+h(X)+Zq(X)$. I don't know how to cancel the $Z$ coming from $Zf_{2}(X,Y)$ with the relations. Other than that, everything is clear (after looking around someone explained why division with remainder works in a general domain using monic polynomials); I probably just need to fiddle around more, I hope. But thank you immensely for your help, Tom! $\endgroup$
    – user68193
    Feb 10, 2014 at 22:39
  • $\begingroup$ Never mind, the argument works with $Yg(X)+h(X)+Zq(X)$ as well, since a polynomial $h(t^3)$ will have powers multiples of 3, and $Y=t^4$ and $Z=t^5$ give powers in $Yg(t^3)$ and $Zq(t^3)$ multiples $1mod3$ and $2mod3$, respectively. Thanks again, Tom! $\endgroup$
    – user68193
    Feb 10, 2014 at 23:08
  • $\begingroup$ @Bleys No problem, happy to help! $\endgroup$ Feb 10, 2014 at 23:10
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    $\begingroup$ This is a good answer, but I think that $X$ has too many meanings. Even though it is clear from context which $X$ is being used, maybe this should be fixed? $\endgroup$
    – john w.
    Feb 16, 2014 at 23:39
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    $\begingroup$ @johnw. I see your point, but I think I'll leave it be. I called the variety $X$ since that was the terminology used in the question, so I'd like to keep that (although I agree, it would probably be better to call it $X'$ or $\bar X$ or something.) I also like to use capital $X$ for variables instead of lowercase $x$ to differentiate the variable from the co-ordinate, which is much more likely to lead to confusion. I don't like using $X_1,X_2,\dots$ for variables if I can avoid it since it makes it harder to understand what's going on at a glance. $\endgroup$ Feb 17, 2014 at 13:55
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What you're looking for is the relation ideal of $t^3$, $t^4$, $t^5$, that is $$\{ f \in k[x,y,z] \mid f(t^3,t^4,t^5) = 0 \}.$$ There are algorithmic ways to compute this: consider the ideal $J = (x - t^3, y - t^4, z - t^5)$ of $k[x,y,z,t]$ and look at the intersection $J \cap k[x,y,z]$; this is what you're looking for and Grobner basis can be used to compute it.

In this case, however, you're probably supposed to just come up with some elements of this relation ideal (e.g., $(t^3)^2 - (t^4)(t^5) = 0$, so $x^2 - yz$ is in the relation ideal) and hope that you've found them all.

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Just a comment (but too long for an actual comment). Over in this thread [number 6) to Georges's answer] it is stated that the image of every morphism (over $k$) $\mathbb{A}^1_k\to\mathbb{A}^n_k$ is closed. This immediately solves the closedness part of this problem (since your set is obviously parametrized by $\mathbb{A}^1$).

To see the claim, consider the induced $k$-algebra map $k[y_1,\ldots,y_n]\to k[x]$. We have two choices: either all the $y_i$ map to constants $c_i\in k$ or the map is finite. If its the former, the image of $\mathbb{A}^1$ is the $k$-rational point $(c_1,\ldots,c_n)$ and if its the latter our map is a finite map, and so closed by the Lying Over Theorem.

Once again, just a comment :)

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  • $\begingroup$ Alternatively, you can compactify to obtain a map $\Bbb P^1_k \rightarrow \Bbb P^n_k$, and deduce the result from closedness of projective morphisms (cf. here). $\endgroup$ Feb 11, 2014 at 14:52

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