We've been given the set $X = \{(t^3,t^4,t^5) \in \mathbb{A}^3 \mid t \in \mathbb{A}^1\}$ (where the underlying field $\mathbb{K}$ is infinite), and have been asked to show that $X = \mathbb{V}(J)$ where $J = \langle xz -y^2, x^3 - yz, z^2 - x^2 y \rangle$, which I have managed to prove. However, it then asks us to show that $\mathbb{I}(X) = J$ i.e. $\mathbb{I(V}(J)) = J$, and I'm unsure on how to approach it. One inclusion is true for any $J$, so it remains to show $\mathbb{I(V}(J)) \subseteq J$. I've tried writing any polynomial in $\mathbb{K}[x,y,z]$ in the form $f = f_1 (xz-y^2) + f_2 (x^3 -yz) + f_3 (z^2 - x^2 y) + g$, aiming to show that if $f \in \mathbb{I(V}(J))$, then $g = 0$, but haven't managed to get anywhere.

Is this the right way to approach it?


Your general approach is sensible, but you may need to be more systematic.

The last relation lets you eliminate all powers of $z$ beyond the first. So working modulo $J$, any coset has a representative of the form $f_1(x,y) + f_2(x,y) z$. Now can you use the first two relations to simplify $f_2$ any further?

  • $\begingroup$ So this is my current line of thinking (correct me if I'm wrong): Write $zf_2(x,y) = z(g(x,y) + h(y))$ where $g$ contains no terms only involving $y$. By the first relation, we can write $zg(x,y)$ as $f_3(y)$ modulo $J$, and by the second relation, we can write $zh(y) = f_4(x)$ modulo $J$. Thus, modulo J, we have $f = f_1(x,y) + f_3(y) + f_4(x) = g(x,y)$. $\endgroup$ – lokodiz Oct 19 '13 at 16:12
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    $\begingroup$ @SimonC: Dear Simon, There is the possibility of having a monomial of $z$, which can't be absorbed into other expressions. (Perhaps in your analysis you overlooked the possibility of a constant term in $h(y)$?) And then there are non-trivial relations between $x$ and $y$ coming from $J$, which you will now want to take into account. Regards, $\endgroup$ – Matt E Oct 19 '13 at 20:48
  • $\begingroup$ yeah, you're right. It's late now (here at least) but have you any further hints? $\endgroup$ – lokodiz Oct 19 '13 at 22:56
  • $\begingroup$ @SimonC: Dear Simon, Well, you're almost done, but you need to find the relation between $x$ and $y$. If you think about the fact that it should be induced by the formulas $x = t^3$, $y = t^4$, you shouldn't have trouble figuring it out. Regards, $\endgroup$ – Matt E Oct 19 '13 at 23:52
  • $\begingroup$ @ Matt E: I know that x^4 = y^3, so should I write $f_i(x,y) = g_1(x) + yg_2(x) + y^2g_3(x)$ modulo $J$? I really don't see where I'm going with this... $\endgroup$ – lokodiz Oct 21 '13 at 8:50

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