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I am wondering if this is true or not: $\mathbb{C}\otimes_\mathbb{Q}\mathbb{C} \cong \mathbb{C}\times\mathbb{C}$ (as rings). Any help or suggestion would be helpful.

Sara
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3 Answers3

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The equation $x^2=2$ has (at least) $8$ distinct solutions $$\pm \sqrt{2} \otimes 1$$ $$1 \otimes \pm \sqrt{2}$$ $$\pm i \sqrt{2} \otimes i$$ $$i \otimes \pm i\sqrt{2}$$ in $\mathbf C \otimes_{\mathbf Q} \mathbf C$, whereas it only has $4$ solutions $(\pm \sqrt 2, \pm \sqrt 2)$ in $\mathbf C\times \mathbf C$.

Bruno Joyal
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  • @Mike101 Because the tensor product is over a field: given bases of $V$ and $W$, there is an explicit basis of $V \otimes W$... – Bruno Joyal Jun 10 '14 at 08:28
  • In fact, $v \otimes w = v' \otimes w'$ for non-zero vectors $v,v',w,w'$ iff $v'=\lambda v$ and $w'=\lambda^{-1} w$ for some invertible scalar $\lambda$. – Martin Brandenburg Jun 10 '14 at 08:34
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The only proper nontrivial quotients of $\Bbb C\times\Bbb C$ are $\cong\Bbb C$ itself.

But $\Bbb C\otimes_{\Bbb R}\Bbb C\not\cong\Bbb C$ is a proper nontrivial quotient of $\Bbb C\otimes_{\Bbb Q}\Bbb C$.

anon
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  • @Mike101 the map $a\otimes b\mapsto a\otimes b$ – anon Jun 10 '14 at 08:19
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    http://math.stackexchange.com/questions/181788 – Martin Brandenburg Jun 10 '14 at 08:30
  • @Mike101 $\Bbb C\otimes_{\Bbb R}\frac{\Bbb R[x]}{(x^2+1)}\cong\frac{\Bbb C[x]}{(x^2+1)}\cong\frac{\Bbb C[x]}{(x+i)}\times\frac{\Bbb C[x]}{(x-i)}\cong\Bbb C\times\Bbb C$ – anon Jun 10 '14 at 08:35
  • $R\otimes_S S[x]\cong R[x]$ for any ring $R$ and subring $S\subseteq R$. Can you think of what the isomorphism is? Hint: the tensor symbol is designed to look like multiplication. – anon Jun 10 '14 at 08:38
  • General properties of tensor products can be found in any introduction to tensor products (and have been asked 100 times on math.SE). These comments here should be about the answer by sea turtles. – Martin Brandenburg Jun 10 '14 at 08:45
  • @Mike101 If you can find the isomorphism for $R\otimes_S S[x]\cong R[x]$ then you can do find it for $R\otimes_SS[x]/(f(x))\cong R[x]/(f(x))$. – anon Jun 10 '14 at 08:56
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$\mathbb{C} \times \mathbb{C}$ is noetherian, but $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ is not noetherian.

Specifically, $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2})} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2},\sqrt{3})} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})} \mathbb{C} \dotsc$ is an infinite sequence of surjective ring homomorphisms which are no isomorphisms.

Community
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Martin Brandenburg
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