Observe that \((y^2+2xy-1) + (x^2+1) = x^2+2xy+y^2=(x+y)^2\). Thus \((x+y)^2\) is in the ideal. However, \(x+y\) is not in the ideal; this can be checked in Sage:

sage: R.<x,y> = PolynomialRing(QQ)
sage: ideal(y^2+2*x*y-1, x^2+1).reduce(x+y)
x + y

Thus the ideal is not radical.

The set \(\mathbf{V}(J)\) is the set of points in \(k^2\) such that \(xy=0\) and \((x-y)x=0\). If \(x \ne 0\), then \(y=0\) and \(x-y=0\) have no common solutions. If \(x=0\) then both equations are satisfied. Thus \(\sqrt{J}=\mathbf{I}(\mathbf{V}(J))\) is the set of all polynomials \(f\) which vanish for all points where \(x=0\). Thus \(f \in \sqrt{J}\) if and only if \(f(0,y)=0\) for all \(y\). Thus \(\sqrt{J}=\langle x \rangle\) as desired.

Suppose \(f \in (I+J)K\). Then \[f = \sum_{s=1}^r g_sk_s\] where \(g_1,\ldots,g_r \in I+J\) and \(k_1, \ldots, k_r \in K\). For each \(s \in \{ 1, \ldots, r\}\) we have \(g_s=i_s+j_s\) for \(i_s \in I\) and \(j_s \in J\) since \(g_s \in I+J\). Thus \[\begin{align*} f &= \sum_{s=1}^r (i_s+j_s)k_s\\ &= \left(\sum_{s=1}^r i_sk_s\right) + \left(\sum_{s=1}^r j_sk_s\right)\\ &\in IK+JK \end{align*}\] Thus we’ve shown that \((I+J)K \subseteq IK+JK\).

For the other direction, suppose \(f \in IK\). Then \[f = \sum_{s=1}^r i_sk_s\] where \(i_1,\ldots,i_r \in I\) and \(k_1, \ldots, k_r \in K\). Since \(I \subseteq I+J\), we have \(i_1,\ldots,i_r \in I+J\). Thus \(f \in (I+J)K\). Thus \(IK \subseteq (I+J)K\). Similarly, one can show that \(JK \subseteq (I+J)K\). Since \(IK+JK\) is the minimal ideal containing \(IK\) and \(JK\), we conclude that \(IK+JK \subseteq (I+J)K\).

Suppose \(f \in \sqrt{I}\). Then \(f^m \in I\) for some integer \(m > 0\). Thus \((f^m)^\ell \in I^\ell\). Thus \(f^{m\ell} \in J\) since \(I^\ell \subseteq J\). Thus \(f \in \sqrt{J}\) as desired.

By definition, \(h \in \langle f \rangle : \langle g \rangle\) if and only if such that \(hk \in \langle f \rangle\) for all \(k \in \langle g \rangle\). Observe that \(k \in \langle g \rangle\) if and only if \(k=gl\) for some \(l \in k[x,y]\). And \(hk \in \langle f \rangle\) if and only if \(f\) divides \(hk\). So \(h \in \langle f \rangle : \langle g \rangle\) if and only if \(f\) divides \(hlg\) for all \(l \in k[x,y]\). This, in turn, is equivalent to \(f\) dividing \(hg\).

Thus, we wish to determine for which \(h \in k[x,y]\) the polynomial \((x+y)^2(x-y)(x+z^2)\) divides \(h(x+z^2)^3(x-y)(z+y)\). This occurs precisely when \((x+y)^2\) divides \(h\).

Thus \(\langle f \rangle : \langle g \rangle = \langle (x+2)^2 \rangle\).

Assume \(I\) is radical. Suppose \(f^m \in I:J\) for some positive integer \(m\). Then \(hf^m \in I\) for every \(h \in J\). Thus \(h^mf^m \in I\) for every \(h \in J\). Since \(I\) is radical, \((hf)^m \in I\) implies that \(hf \in I\). Thus \(hf \in I\) for every \(h \in J\). Thus \(f \in I:J\).

Suppose \(IJ \subseteq K\) and \(f \in I\). Since \(IJ \subseteq K\), we have \(fg \in K\) for all \(g \in J\). Thus \(f \in K:J\). Thus \(I \subseteq K:J\).

Now suppose \(I \subseteq K:J\). Suppose \(a \in I\) and \(b \in J\). Since \(a \in I\) we have \(a \in K:J\). Thus \(ab \in K\) since \(b \in J\). Thus \(ab \in K\) whenever \(a \in I\) and \(b \in J\). Now suppose \(f \in IJ\). Then \(f = \sum_{i=0}^s a_ib_i\) for \(a_1, \ldots, a_s \in I\) and \(b_1, \ldots, b_s \in J\). By the above \(a_1b_1, \ldots, a_sb_s\) are all in \(K\). Since \(K\) is an ideal and is closed under sums, \(f \in K\). Thus \(IJ \subseteq K\).

Let \(I\) be a prime ideal in \(k[x_1,\ldots,x_n]\). Suppose \(f^m \in I\) for some \(f \in k[x_1,\ldots,x_n]\). Let \(r\) be the smallest positive integer such that \(f^r \in I\). Now \(f \cdot f^{r-1}=f^r \in I\), so either \(f \in I\) or \(f^{r-1} \in I\) since \(I\) is prime. If \(f \notin I\), then \(r > 1\) and \(f^{r-1} \in I\); but this contradicts the choice of \(r\). Thus we conclude \(f \in I\). Since we only assumed \(f^m \in I\), we have shown that \(I\) is radical.