1
$\begingroup$

Let $Mat_2(\mathbb{Z})$ be the $\mathbb{Z}$-algebra of $2\times2$ matrices with integer entries. Let $A$ be a $\mathbb{Z}$-submodule of $Mat_2(\mathbb{Z})$ containing $\mathbb{Z}$. We want to show that $A$ is a subring, does it suffice to show that the product of two elements in $A$ lies in $A\otimes\overline{\mathbb{Q}}$?

Seen that if $A$ is the $\mathbb{Z}$-submodule $\mathbb{Z}\oplus 2\mathbb{Z}\oplus0\oplus0$ of $B=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$, the equality $(A\otimes\overline{\mathbb{Q}})\cap B=A$ does not hold.

I do not understand how does it work the argument of the notes of B. Conrad http://math.stanford.edu/~conrad/676Page/handouts/picgroup.pdf page 4, the 14th-16th lines of section 3.

Improve this question
$\endgroup$
2
  • $\begingroup$ I don't know the answer to your first question, but you don't need it to understand Brian Conrad's argument. In his setting, $A$ is a submodule of $\mathrm{End}_{\mathbb Z}\left(M\right)$ determined by the single equality $f^{\dagger} = \mathrm{Tr}\left(f\right) \cdot 1_M - f$ (you can forget the "$\in \mathrm{End}_{\mathbf Z}\left(M\right)$" part, because that follows automatically from the equality), and this equality clearly preserves its meaning upon base extension. $\endgroup$
    – darij grinberg
    Jan 15, 2014 at 17:33
  • $\begingroup$ OK, the answer to your first question is actually "No". For a counterexample, take the $\mathbb Z$-module of all $2\times 2$-matrices $A \in \mathrm{Mat}_2\left(\mathbb Z\right)$ such that the sum of the two entries of $A$ not on the main diagonal is even. $\endgroup$
    – darij grinberg
    Jan 15, 2014 at 17:35

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.