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Let $\mathbb{Z}[i]=\left\{a+bi:a,b \in \mathbb{Z}\right\}$ be the ring of Gaussian integers $(i^{2}=-1)$ and let $\mathbb{Q}[i]=\left\{a+bi:a,b \in \mathbb{Q}\right\}$ be its fraction field.

1) Show that the tensor product $\mathbb{Q}[i] \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ is isomorphic to the product $\mathbb{Q}[i] \times \mathbb{Q}[i]$ of two copies of the field.

2) Let $S=\left\{1,2,2^2,2^3,...\right\} \subseteq \mathbb{Z}$ be the multiplicatively closed subset generated by 2. Find two distinct idempotents in the algebra $S^{-1}(\mathbb{Z}[i] \otimes_{\mathbb{Z}} \mathbb{Z}[i])$ other than $0$ and $1$.

For the first part, I thought I would try to use the presentation $\mathbb{Q}[i] \simeq \mathbb{Q}[T]/(T^2+1)$ and see what comes from that. Any help would be much appreciated!

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    $\begingroup$ By $\mathbb Z(i)$, you mean $\mathbb Z\left[i\right]$ ? $\endgroup$ – darij grinberg Oct 7 '13 at 1:33
  • $\begingroup$ I agree with Darij, I doubt you meant parentheses. As for 1) you can use the fact that if $R$ is an $A$-algebra then $R\otimes_A A[t]/(f(t))\cong R[t]/(f(t))$. $\endgroup$ – Alex Youcis Oct 7 '13 at 2:31
  • $\begingroup$ Ah yes, you are right! I'll change it. $\endgroup$ – user 3462 Oct 7 '13 at 4:00
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2) In this topic you can find that $\mathbb Z[i]\otimes_{\mathbb Z}\mathbb Z[i]\cong\mathbb Z[i][X]/(X^2+1)$ and this implies that $S^{-1}(\mathbb Z[i]\otimes_{\mathbb Z}\mathbb Z[i])\cong(S^{-1}\mathbb Z)[i][X]/(X^2+1)=\mathbb Z[\frac 12,i][X]/(X^2+1)$. Following the same calculations you get that $\frac 12\pm(\frac 12i)x$ are the nontrivial idempotents.

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Hints:

1) Your idea of using the isomorphism $\Bbb{Q}[i]\cong\Bbb{Q}[T]/\langle T^2+1\rangle$ is a good one. General facts coming to fore here are the following. If $K$ is an extension field of $\Bbb{Q}$ and $p(T)\in\Bbb{Q}[T]$ is a polynomial, then we have an isomorphism $$ K[T]/\langle p(T)\rangle\cong K\otimes_{\Bbb{Q}}\left(\Bbb{Q}[T]/\langle p(T)\rangle\right). $$ This is because $K$ is flat as a $\Bbb{Q}$-module, and we have a short exact sequence $$ 0\to \Bbb{Q}[T]\to \Bbb{Q}[T]\to \Bbb{Q}[T]/\langle p(T)\rangle\to0, $$ where the inclusion is multiplication by $p(T)$.

Another fact that you need: If $p(T)\in K[T]$ is a product of coprime polynomials, $p(T)=f(T) g(T)$, then the Chinese Remainder Theorem tells us that $$ K[T]/\langle p(T)\rangle\cong \left(K[T]/\langle f(T)\rangle\right)\oplus \left(K[T]/\langle g(T)\rangle\right). $$

2) The multiplicative identities of the fields that appear as summands of the tensor product in part 1 (i.e. $(1,0)$ and $(0,1)$) are clearly idempotents different from both zero and one. Can you show that their images under the inverse mapping of the above CRT-isomorphism already reside in the localization w.r.t. to the multiplicative set of powers of two?


After the dust has settled down you should have found the idempotents $$ e_{\pm}=\frac12(1\otimes1\pm i\otimes i). $$

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