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I am stuck on some proofs concerning strong approximation in Chapter 3.1 of Hida's book on modular forms. I have put in green the things that I do not understand.

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The set $gL\subset \mathbb{A}^\infty$ is a (free) module over $\widehat{\mathbb{Z}}$. But why can we assume that its basis vectors are in $\mathbb{Q}^n$? I think that it has something to do with the identity $\mathbb{A}^\infty=\mathbb{Q}+\widehat{\mathbb{Z}}$, but I do not see in what way?

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We have $s,g\in GL_n(\mathbb{A}^\infty)$ and $X\in GL_n(\mathbb{Q})$ (why again? this has to do with the previous proof). Then $s^{-1}g^{-1}X$ surely is in $SL_n(\mathbb{A}^\infty)$, but why is it suddenly in $SL_n(\widehat{\mathbb{Z}})$?

Any help is much appreciated.

1 Answers1

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For your first question about a rational $\hat{\Bbb{Z}}$-basis of $gL$.

Take $m\in \Bbb{Z}-0$ such that $mg$ is in $M_n(\hat{\Bbb{Z}})$. Let $h = Adj(mg)\in M_n(\hat{\Bbb{Z}})$.

Then $gh = cI$ with $c=m^{n-1}\det(g)\in (\Bbb{A}^{(\infty)})^\times$.

So $gL$ contains $c L$.

$cL=c'L$ with $c'\in \Bbb{Q}^\times$.

$c'L$ has finite index in $gL$ and the $J$ elements of $gL/c'L$ have representatives in $\Bbb{Q}^n$.

Therefore $$gL = c'L + \sum_{j=1}^J v_j \hat{\Bbb{Z}} = (\sum_{i=1}^n e_i c' \Bbb{Z}+\sum_{j=1}^J v_j \Bbb{Z})\otimes_\Bbb{Z} \hat{\Bbb{Z}} = (\sum_{i=1}^n w_i \Bbb{Z})\otimes_\Bbb{Z} \hat{\Bbb{Z}}=\sum_{i=1}^n w_i \hat{\Bbb{Z}}$$

reuns
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  • Thank you for your answer! Could you please elaborate on the sentence after $cL=c'L$ and explain your notation in the last sentence ? – Dr. Heinz Doofenshmirtz Dec 12 '21 at 11:22
  • $gL$ is generated by $c'$ times the canonical basis (the $c' e_i$) plus finitely many elements (the $v_j$) and they all are in $\Bbb{Q}^n$. Take a free basis of the $\Bbb{Z}$-module they generate, it will give a free $\hat{\Bbb{Z}}$ basis of $gL$. @Dr.HeinzDoofenshmirtz – reuns Dec 12 '21 at 17:13