8

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?

Pooya
  • 81

1 Answers1

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)

  • 1
    $(x+y)$ 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 08:03
  • 4
    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 09:22
  • @Pooya Well, it is not possible to completely characterise prime ideals of the tensor products in terms of prime ideals of each of the factors. However, this might help. From what I understand, you can think of $\operatorname{Spec} (A\otimes B)$ as small collections of ideals indexed by pairs $(p,q)$, $p\in \operatorname{Spec} A$, $q\in \operatorname{Spec} B$, corresponding to $\operatorname{Spec} (k(p)\otimes k(q))$. – Dog_69 Dec 14 '21 at 14:23