Showing tensor product of f.g. reduced $k$-algebras is reduced where I should assume $k$ algebraically closed? This is a rough idea I have to prove the statement of the following. I want to prove algebraically the product of varieties is variety where variety means irreducible algebraic set.

Let $A,B$ be f.g. reduced $k$-algebras and I should assume $k$ algebraically closed. Then $A\otimes B$ is reduced.

From Noether normalization, I obtain $k[x_1,\dots, x_n]\to A$ as integral extension and $k[y_1,\dots, y_m]\to B$ as integral extension. And Noether normalization does not require $k$ algebraically closed. So I obtained the following natural map. Denote $R_1=k[x_1,\dots, x_n],R_2=k[y_1,\dots, y_m]$.
$R=R_1\otimes R_2=k[x_1,\dots, x_n,y_1,\dots, y_m]\to A\otimes B$. It suffices to show any $0\neq a\otimes b\in A\otimes B$ is integral over $R$. Clearly $a\otimes 1$ integral over $R$ and $1\otimes b$ integral over $R$ as well. So $R\to R[a\otimes 1]\to R[a\otimes 1,1\otimes b]$ are the extensions and both extensions are finite. So I conclude that $a\otimes b$ is integral over $R$. So if I have $0\neq x,y\in A\otimes B$ and $xy=0$ say $x$ satisfies equation $f(z)=0$ and $y$ satisfies $g(z)=0$ where $f,g\in R[z]$ are monics. Choose $f,g$ containing constant term say $f_0,g_0\in R$. Then I check $f(x)g(y)=0$ by plugging in $x$ and $y$ values correspondingly. However $xy=0$ implies $f_0g_0=0$. Since $R=k[x_1,\dots, x_n,y_1,\dots, y_n]$ is integral domain, I conclude that $g_0=0$ or $f_0=0$ which is contradiction. So $xy\neq 0$ for any $0\neq x,y\in R$.


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*I did not use that $k$ is algebraically closed. I feel I have done something wrong here.


*I think one the possible mistake would be $R_1\otimes R_2\to R_1\otimes B\to A\otimes B$'s last extension as there is no reason to believe $B$ is flat over $k$ as a module first.

 A: Yes your point (2) looks valid to me but that's not the only thing that is going wrong. 
Indeed, you can have a case where the inclusion $R_1 \otimes R_2 \to A \otimes B$ is an injective inclusion which is also integral but the theorem does not hold.
The mistake lies where you choose $f,g$ with constant terms $f_0, g_0$. You may not be able to do that for example if $f = Z^m$. Even otherwise when you plug $x$ into $f$, the expression is defined in a ring which may not be an integral domain (that's what you are trying to prove), so you cannot cancel off the powers and make that assumption. 
I will give an example which shows that it may hold that the extension is injective and integral but your result is not true because of the aforementioned point. 
Take $k = \mathbb F_p(t)$. $A = B = \mathbb F_p(t^{1/p})$. (This is a $k$ algebra)
Now you can take $R_1 = R_2= k$ because the extension $k \to \mathbb F_p(t^{1/p})$ is integral. Now do the tensor products and you get an extension 
$\mathbb F_p(t) \to \mathbb F_p(t^{1/p}) [X]/ (X - t^{1/p})^p$. This is injective  and integral.
But clearly this is not reduced. Also the polynomial annihilator for $X - t^{1/p}$ in $\mathbb F_p(t)$ is $Z^p$. 
