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Let $R$ and $S$ be commutative rings over a field $k$. Let $I$ be an ideal of the tensor ring $R\otimes_{k} S$. It is true that there exist ideals $I_{1}$ and $I_{2}$ of $R$ and $S$ respectively such that $$ I=I_{1}\otimes_{k} I_{2}? $$ If this is not true, are there any description of $I$? What if we don't assume commutativity of one of rings?

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The simplest example of the setup is probably

  • $R = k[x]$
  • $S = k[y]$
  • $R \otimes_k S = k[x,y]$

and the simplest ideals of $R \otimes_k S$ are principal ideals. The first place to look for such a thing that is a counter-example would be to pick a generator that isn't obviously a product of something from $k[x]$ and something from $k[y]$.

(P.S. I think you can arrange for an $I_1$ and an $I_2$ such that the "inclusion" from $I_1 \otimes_k I_2$ to $R \otimes_k S$ isn't monic, so really it isn't an ideal)

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$(x+y)$ works. I am sorry that this is too easy question. I should have thought more. This example shows that there is no good description of ideals of $R\otimes S$, I think. – Pooya Sep 26 '12 at 8:03
Dear Hurkyl, the morphism $I_1 \otimes_k I_2 \to R \otimes_k S$ is injective because all modules over a field are flat. – Georges Elencwajg Sep 26 '12 at 9:22

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