Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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)

share|improve this answer
    
$(x+y)$ 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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.