8
$\begingroup$

I wanted to verify that for a connected ring $k$ (i.e. a ring with connected spectrum, or equivalently without idempotents other than $0$ and $1$) the group of $k$-endomorphisms of the multiplicative group $\mathbf{G}_m$ over $k$ can be identified with $\mathbf{Z}$. I have an outline for this (at least for $k$ local) which suggests that I treat the case of a field first, then extend it to Artin local rings by induction on length, then to complete Noetherian local rings, etc.. But the very simple argument I have for the case of a field seems to me to be completely general, which makes me believe I must be making a mistake or overlooking something.

Using the definition of the multiplication for $\mathbf{G}_m$, it can be shown that $f(t)\in k[t,t^{-1}]^\times$ ($k[t,t^{-1}]$ being the coordinate ring of $\mathbf{G}_m$) yields a $k$-homomorphism if and only if $f(xy)=f(x)f(y)$ in $k[x,y,x^{-1},y^{-1}]$. Note that $\mathbf{G}_m(\mathbf{G}_m)=\mathrm{Hom}_{k-\mathrm{Alg}}(k[t,t^{-1}],k[t,t^{-1}])=k[t,t^{-1}]^\times$ via $\varphi\mapsto\varphi(t)$, so $f$ must be a unit to have a $k$-morphism at all.

Now, if I write $f=\sum_n a_n t^n$, $n$ ranging over all of $\mathbf{Z}$, then the equation $f(xy)=f(x)f(y)$ seems to give me

$\sum_n a_n x^n y^n=\sum_{n,m} a_n a_mx^n y^m$.

Because $f$ is a unit, and in particular non-zero, some $a_n$ must be non-zero, say $a_{n_0}\neq 0$. Equating coefficients of monomials in the equation above (using that the monomials $x^n y^m$ for $n,m\in\mathbf{Z}$ form a $k$-basis for $k[x,y,x^{-1},y^{-1}]$) we get $a_n^2=a_n$ for all $n$ and $a_na_m=0$ for $n\neq m$. Since $k$ is connected and $a_{n_0}\neq 0$, $a_{n_0}^2=a_{n_0}$ forces $a_{n_0}=1$. Then, for any $m\neq n_0$, $0=a_{n_0}a_m=a_m$, so $f(t)=t^{n_0}$.

Is there anything wrong with this argument? Have I used somewhere that $k$ is a field?

Incidentally, it also seems like the equation $f(xy)=f(x)f(y)$ for $f\neq 0$ implies that $f$ is a unit, since all I used in deducing that $f(t)=t^{n_0}$ for some $n_0$ was that $f$ was non-zero.

$\endgroup$
  • 1
    $\begingroup$ I don't see any flaw here. $\endgroup$ – user18119 Jul 27 '13 at 5:29
  • $\begingroup$ Dear @QiL'8, Thank you for your comment. I feel better about this knowing that it looks correct to someone with more experience in the subject. $\endgroup$ – Keenan Kidwell Jul 27 '13 at 16:54
  • $\begingroup$ I agree, the proof is OK. $\endgroup$ – Martin Brandenburg Aug 5 '13 at 17:58
  • $\begingroup$ Thanks @Martin. I was just alarmed because I was expecting to have to do a lot more work to get the result for general local (or connected) rings. $\endgroup$ – Keenan Kidwell Aug 5 '13 at 18:13
2
$\begingroup$

QiL'8 and Martin Brandenburg have passed judgement on the proof, so in the interest of answering the question, I'm posting this as an answer. It seems the proof is correct.

$\endgroup$

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.