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!

  • 1
    $\begingroup$ 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

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.



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 $$ e_{\pm}=\frac12(1\otimes1\pm i\otimes i). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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