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I'm reading this blog post about $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ and I have two questions about it:

  1. Is a simple tensor a tensor that cannot be written as a sum of tensors?

  2. On the first line, how did they get $\frac{ad}{bd} \otimes \frac{bc}{bd} = \frac{a}{b} d \otimes b \frac{c}{d}$ ?

Many thanks for your help.

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Matt: Compare with Theorem 4.17 and the paragraph after Remark 4.18 in the other tensor product notes you are reading. –  KCd Oct 13 '11 at 15:46
    
@KCd: thank you, will do! –  Matt N. Oct 13 '11 at 16:03
    
@KCd On page 1 here in the penultimate paragraph you wrote "finite free modules". Do they really have to be finite to define equality? I think they could be infinite and free and we could define equality but I think I'm missing something. –  Matt N. Apr 23 '12 at 20:16

1 Answer 1

up vote 7 down vote accepted

A simple tensor is an element of a tensor product that can be written in the form x⊗y. It could also be written as a sum of tensors, but not all sums of tensors can be written as a single x⊗y (in general).

The chain of equalities in the blog post was incorrect. They are not equal. Here is a similar but correct chain of equalities you might find useful:

$$ \frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{c}{d} = \frac{a}{bd} d \otimes c \frac{1}{d} = \frac{a}{bd} c \otimes d \frac{1}{d} = \frac{ac}{bd} \otimes 1$$

I would be hesitant to write $\frac{ac}{bd}(1\otimes 1)$ unless you are considering $\mathbb{Q}$ as a left-$\mathbb{Q}$, right-$\mathbb{Z}$ module.

At any rate, the map $q \mapsto q \otimes 1$ is an isomorphism from $\mathbb{Q}$ to $\mathbb{Q}\otimes \mathbb{Q}$, or indeed from $\mathbb{Q}$ to $\mathbb{Q} \otimes N$ for any abelian group $\mathbb{Z} \leq N \leq \mathbb{Q}$.

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So just to make sure I understand this: $a \otimes b + c \otimes d$ is also a simple tensor? –  Matt N. Oct 13 '11 at 14:24
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I'm the author of the post you linked. I just wanted to say thanks for pointing out my mistake. The post has now been edited. –  user17581 Oct 13 '11 at 14:54
    
Dear Jack: +1! For the last sentence of your answer, isn't it sufficient that $N$ be nonzero? –  Pierre-Yves Gaillard Oct 13 '11 at 14:59
    
@Pierre-Yves No. Take any abelian group which is torsion as a $\mathbb{Z}$-module. Tensoring with $\mathbb{Q}$ gives $0$. E.g. $N=\mathbb{Z}/5\mathbb{Z}$ gives $\mathbb{Q}\otimes N=0$ –  Matt Oct 13 '11 at 16:14
    
@Matt: in Q⊗Q, yes, every tensor is simple, but in (Q⊕Q)⊗(Q⊕Q) there are tensors that cannot be written in the form x⊗y. If you believe Q^n ⊗ Q^n is the space of n×n matrices over Q, then (nonzero) simple tensors are the matrices of rank 1. –  Jack Schmidt Oct 13 '11 at 16:29

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