Let $K$ be an algebraically closed field, and let $I$ be an ideal of $K[x_1,\dots,x_n]$ ($n \geq 2$). We suppose that for all $i$, and for all $a_1,\dots,a_{i1},a_{i+1},\dots,a_n \in K^{n1}$, the image of $I$ in the specialization $K[x_1,\dots,x_n]/(x_j = a_j, j \neq i) \simeq K[x_i]$ is a radical ideal. Is it then true that $I$ itself is radical?

3$\begingroup$ That is already false for $n=2$ and the ideal $I=\langle x_1x_2,x_1^2,x_1x_2,x_2^2\rangle$. For every choice of $a_1$, resp. of $a_2$, the image of $I$ is a radical ideal in $K[x_2]$, resp. in $K[x_1]$. $\endgroup$– Jason StarrJul 11 '16 at 18:59

$\begingroup$ Intuitive explanation of the counterexample posted by Jason Starr: imagine a point thickened in the direction of the main diagonal (of course, $I=\langle x_1x_2,x_1^2\rangle=\langle x_1x_2,x_2^2\rangle$). Intersecting it with any line transverse to this diagonal just gives a (reduced) point. $\endgroup$– GroTsenJul 11 '16 at 19:06

1$\begingroup$ Incidentally, the converse is also false: $\langle (x_1x_2)(x_1+x_2)\rangle$ is radical, but specializing to $x_1=0$ or to $x_2=0$ gives a thickened point. $\endgroup$– GroTsenJul 11 '16 at 19:10
I am just posting the comment above as an answer. The ideal $I$ is not always radical. One counterexample is $I=\langle x_1  x_2, x_1^2,x_1x_2,x_2^2 \rangle$. As GroTsen points out (and this is also how I think of the example), this is the ideal cutting out a nonreduced scheme of length $2$ with support at the origin of the affine plane and with tangent vector in the direction of the diagonal. Since that direction is neither the direction of the $x$axis nor the $y$axis, intersection with those two axes does not detect the nonreducedness of the scheme.