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Problem of the Week #266 - Feb 27, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Let $\Bbb Z$ denote the ring of Gaussian integers. Prove that the tensor product $\Bbb Z\otimes_{\Bbb Z} \Bbb R$ is ring isomorphic to the complex numbers $\Bbb C$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
I'm giving users one more week to attempt to solve this problem. Start by considering simple tensors $1 \otimes x$ and $i \otimes x$ for a fixed $x\in \Bbb R$.
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.

The map $\Bbb Z \times \Bbb R \to \Bbb C$, $(z,r) \mapsto zr$ is a $\Bbb Z$-bilinear map, so it induces a unique $\Bbb Z$-linear map $\Phi : \Bbb Z\otimes_{\Bbb Z} \Bbb R \to \Bbb C$ such that $\Phi(z\otimes r) = zr$. The map $\Psi: \Bbb C \to \Bbb Z\otimes_{\Bbb Z} \Bbb R$ given by $\Psi(z) = 1\otimes \operatorname{Re}(z) + i \otimes \operatorname{Im}(z)$ is the inverse of $\Phi$. Hence, $\Phi$ is a $\Bbb Z$-linear isomorphism. So it suffices to show $\Psi(zw) = \Psi(z)\Psi(w)$ for all $z,w\in \Bbb C$. Given $z = a + bi$ and $w = c + di$, $zw = (ac - bd) + (ad + bc)i$; thus $$\Psi(zw) = 1\otimes (ac - bd) + i\otimes (ad + bc) = (1\otimes a)(1\otimes c) + (i\otimes b)(i\otimes d) + (1\otimes a)(i\otimes d) + (i\otimes b)(1\otimes c) = (1 \otimes a + i\otimes b)(1\otimes c + i\otimes d) = \Psi(z)\Psi(w).$$
 
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