Find $f \in I(V(J))$ but $f \notin J$

Question

Let $A := K[x,y,z]$ be a polynomial ring and consider the following ideals of $A$:

1. $(y^2 - x^4, x^2 - 2x^3 - x^2y + 2xy + y^2 - y)$,
2. $(xy + xz + yz, xyz)$,
3. $((x-z)(x-y)(x-2z), x^2 - y^2z)$.

The problem is, for each ideal $J$ defined above, determine polynomials satisfying $f \in I(V(J))$ but $f \notin J$.

Assuming $k$ algebraically closed, we can put $I(V(J)) = \mathrm{Rad}(J)$, so in order to know $I(V(J))$, we only need to know the minimal primes over $J$. That is, minimal elements of the poset $$P := \{\mathfrak{p} \in \mathrm{Spec}(A) : J \subseteq \mathfrak{p}\}$$

Is there an easy way to find such $f$'s without using the strong Nullstellensatz? For the case 2, I've found that the radical is $$(x,y)\cap (x,z) \cap (y,z).$$ However, how can I find such $f$? And how can I solve the other cases?

Answer 1

Assuming $K$ algebraically closed, it is easier to work with closed points of the corresponding varieties.

Take 1. as example. We may determine $V(J)$ as follows:

\begin{eqnarray*} V(J) &=& \{(x, y)\in K^2: y^2 = x^4, (2x + y - 1)(y - x^2) = 0\}\\ &=& \{(x, y)\in K^2: y = x^2\} \cup\{(x, y): y = -x^2, 2x + y - 1 = 0\}\\ &=& \{(x, y)\in K^2: y = x^2\} \cup \{(1, -1)\}. \end{eqnarray*}

Therefore it suffices to find the ideal $I(V(J))$.

If $char(K)= 2$, then the set $V(J)$ becomes $V(y - x^2)$ and it's clear that, since the polynomial $y - x^2$ is irreducible in $K[x, y]$, the ideal $I(V(J))$ is just the principal ideal $(y - x^2)$.

If $char(K)\neq 2$, then we may use the fact $I(X \cup Y) = I(X) \cap I(Y)$ and find that $I(V(J))$ is all the polynomials of the form $(y - x^2)f$ with $f(1, -1) = 0$, which means that $f$ is in the ideal $(x - 1, y + 1)$. Therefore $I(V(J)) = ((y - x^2)(x - 1), (y - x^2)(y + 1))$ in this case.

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As an explicit choice of $f$, we can take e.g. $f = (y - x^2)(x - 1)$. To prove that $f$ is not in the original ideal $J$, one essentially argues in the local ring. An elementary version could be as follows.

Suppose $f$ is in $J$. Then there are polynomials $u, v$ such that $(y - x^2)(x - 1) = (y - x^2)(y + x^2)u + (y - x^2)(2x + y - 1)v$, which means $x - 1 = (y + x^2)u + (2x + y - 1)v$.

This however is not possible: by setting $y = 1 - 2x$, the left hand side is $x - 1$ and the right hand side is a multiple of $(x - 1)^2$.