Ideal of product of affine varieties Let $X\in\mathbb{A}^n$ and $Y\in\mathbb{A}^m$ be affine varieties (irreducible algebraic sets). Then, if we denote by $i_X$ and $i_Y$ their ideals in their respective affine spaces, we define their product in $\mathbb{A}^{n+m}$ as the set
$$
X\times Y = \{(x,y):x\in X\text{ and } y\in Y\} = \{(x,y):x\in Z(i_X)\text{ and }y\in Z(i_Y)\}.
$$
Using the second characterization it's easy to show that in fact $X\times Y$ is closed in $\mathbb{A}^{n+m}$ and particularly that $X\times Y = Z(I_X+I_Y)$, where $I_X$ and $I_Y$ are the vanishing ideals of $X\times\mathbb{A}^m$ and $\mathbb{A}^n\times Y$ (these ideals clearly are in bijective correspondence with $i_X$ and $i_Y$, respectively).
Then, by the Nullstellensatz, we get $I(X\times Y) = \sqrt{I_X+I_Y}$. The problem is that the references I've seen claim that in fact $I(X\times Y) = I_X+I_Y$ (Gathmann's notes, example 2.3.9 in page 25, and Georges' answer in this math.se post, specially the last display). 
This would only be true if $I_X+I_Y$ is radical, which I haven't been able to show for this case and know isn't true in general (see this math.se post). Are my references wrong, or what am I missing?
 A: Lemma. Let $k$ be a perfect field, and let $J_X \subseteq k[x_1,\ldots,x_n]$ and $J_Y \subseteq k[y_1,\ldots,y_m]$ be radical ideals. Denote by $I_X, I_Y \subseteq k[x_1,\ldots,x_n,y_1,\ldots,y_m]$ the ideals they generate in the bigger ring $k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. Then $I = I_X + I_Y$ is radical.
Proof. Let $A = k[x_1,\ldots,x_n]/J_X$ and $B = k[y_1,\ldots,y_m]/J_Y$. The ideal $I$ is the kernel of the ring homomorphism
\begin{align*}
\phi \colon k[x_1,\ldots,x_n,y_1,\ldots,y_m] = k[x_1,\ldots,x_n] \otimes_k k[y_1,\ldots,y_m] &\to A \otimes_k B
\end{align*}
induced by the quotient maps $k[x_1,\ldots,x_n] \to A$ and $k[y_1,\ldots,y_m] \to B$. Moreover, $\phi$ is surjective, so we get an isomorphism
$$k[x_1,\ldots,y_m]/I \stackrel \sim \longrightarrow A \otimes_k B.$$
Since $J_X$ and $J_Y$ are radical, the $k$-algebras $A$ and $B$ are reduced. Over a perfect field, this implies that $A \otimes_k B$ is reduced (see for example this answer I gave a few weeks ago, or see Tag 034N). Thus, $I$ is radical. $\square$
In particular, this settles the question over an algebraically closed field. On the other hand, as Mohan noted, over $k = \mathbb F_p(t)$, you can take $J_X = (x^p - t)$ and $J_Y = (y^p - t)$; then $I = (x^p - t, y^p - t)$, which contains $x^p - y^p = (x-y)^p$, but not $x-y$.
