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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?

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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?

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  • $\begingroup$ 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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