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?
Ideals of the tensor product $R\otimes_{k} S$?
6
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commutative-algebra
tensor-products
1 Answers
3
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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3Dear Hurkyl, the morphism $I_1 \otimes_k I_2 \to R \otimes_k S$ is injective because all modules over a field are flat. – 2012-09-26