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Let $\mathfrak{a}:=(f_1,\ldots,f_r)\subset k[x_1,\ldots,x_n],\mathfrak{b}:=(g_1,\ldots,g_s)\subset k[y_1,\ldots,y_m]$ be radical ideals. Then I wish to prove that $\mathfrak{c}:=(f_1,\ldots,f_r,g_1,\ldots,g_s)\subset k[x_1,\ldots,x_n,y_1,\ldots,y_m]$ is a radical ideal. If we set $x=(x_i)_i,\ y=(y_i)_i$ for simplicity, then otherwise stated, I wish to prove that $$\sqrt{k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}}=k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}=\mathfrak{c}$$ Trivially we have $\supseteq$. Suppose $f^t\in k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}$ for some $t\in\mathbb{N}$, so let $f^t=f'(x,y)f''(x)+g'(x,y)g''(y),\ f',g'\in k[x,y],\ f''\in\mathfrak{a},\ g''\in\mathfrak{b}$. However, I can't figure out the next step from here.

Edit: $k$ is algebraically closed.

Dominic Wynter
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    This is wrong if $k$ is not a perfect field. Basically what you are asking is if $A,B$ are two reduced $k$-algebras, is $A\otimes_k B$ reduced? (can you see why this is simply a translation of your question?) This question has already been asked: http://math.stackexchange.com/q/308015/191425 , and the answer is no in general. Look at Georges Elencwajg's answer. – Hamed Jul 01 '16 at 04:20
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    So this holds for algebraically closed $k$? – Dominic Wynter Jul 01 '16 at 04:31
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    Yes it does, the proof however is not easy. Look up Theorem 3 of Bourbaki's Algebra, Chapter V, §15, as Georges points out... – Hamed Jul 01 '16 at 04:32
  • This seems to have been taken for granted in my textbook on Algebraic Geometry... – Dominic Wynter Jul 01 '16 at 04:34
  • In algebraic geometry books which deal with algebraic varieties, the authors usually say in the beginning (in a hidden comment somewhere) that "unless otherwise stated, all fields are algebraically closed" because that's the setup of algebraic varieties (for example most forms of nullstellensatz are for algebraically closed fields). So I'm not surprised it is taken for granted. – Hamed Jul 01 '16 at 04:38
  • I'm presuming $A=k[x]/\mathfrak{a}$, $B=k[y]/\mathfrak{b}$, then you're saying that $k[x]/\mathfrak{a}\otimes_k k[y]/\mathfrak{b}\cong k[x,y]/(\mathfrak{a}+\mathfrak{b})$? – Dominic Wynter Jul 01 '16 at 04:48
  • Yes, but instead of $\mathfrak{a}+\mathfrak{b}$ it is better to write $(f_1(x), \cdots, f_n(x), g_1(y), \cdots, g_m(y))$ because $\mathfrak{a}, \mathfrak{b}$ are not ideals of $k[x,y]$, rather their extension is. It is just notation though, as long as you know what you're doing, what you said is correct. – Hamed Jul 01 '16 at 04:58
  • By the way this is exactly how one defines the coordinate ring of the product of two affine varieties... – Hamed Jul 01 '16 at 05:00

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