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Let $A$ be an order, i.e. a commutative ring of which the additive group is isomorphic to $\mathbb{Z}^n$ for a certain non-negative integer $n$. Show that there exists an embedding $$A^{\times}_{\text{tor}}\ \hookrightarrow\ (A_{\text{red}})^{\times}_{\text{tor}},$$ where $A^{\times}_{\text{tor}}$ is the group of torsion units of $A$, and $A_{\text{red}}=A/\sqrt{0_A}$ is the reduced ring of $A$.

Edit: In response to a reply which seems to have been removed; I understand that the quotient map $A\rightarrow A_{\text{red}}$ restricts to a group homomorphism $A^{\times}_{\text{tor}}\rightarrow(A_{\text{red}})^{\times}_{\text{tor}}$, but I am unable to show that this map is injective.

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    You're right, I forget one needs to show the quotient map is injective!2012-06-24

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Since this is homework, let me give a hint:

If $a$ lies in the kernel of your map, show that you may write $a = 1 + x$ for some nilpotent $x$. Now write $a^n = 1$ for some $n$, deduce a corresponding equation involving $x$, and see where it leads.

Added: Since the OP has now solved the question, let me sketch the answer, based on the discussion in the comments:

$(1+x)^n = 1$ implies that $nx + $ higher order terms in $x = 0$. From this it is easy to deduce that $x = 0,$ given that $x$ is nilpotent. (The OP's gives the following very succinct approach: we may factor out $x$ in the above equation to get $x (n + $ terms involving positive powers of $x) = 0$, and the parenthetical factor is a non-zero divisor in $A$ (since $n$ is not a zero-divisor, and $x$ is nilpotent).)

As an aside, note that the assumption that $A$ is torsion-free as an abelian group (this is what is really used; of course it follows directly from the assumption that $A$ is an order) is crucial. There are char. $p$ examples where the given map is not injective. One of the simplest is obtained by taking $A = \mathbb F_p[x]/(x^p).$

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    I'm sorry to say that I don't see where it leads.2012-06-25
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    @user: Dear user, What did you try? Regards,2012-06-26
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    If $a$ is in the kernel of this map, then $a=1+x$ for some $x\in\sqrt{0}_A$, and it is a torsion unit. Hence there exist $m,n\in\mathbb{Z}_{>0}$ such that $x^m=0$ and $a^n=1$. The latter yields $$(1+x)^n=x^n+nx^{n-1}+\ldots+nx+1=1,$$ so $x^n+nx^{n-1}+\ldots+nx=0$, and we may reduce this to $$\sum_{i=1}^k{n\choose i}x^i=0,$$ where $k=\min\{m-1,n\}$. On the other hand, we may also write $$nx^{n-1}+n\frac{n-1}{2}x^{n-2}+\ldots+nx=-x^n,$$ which shows that $x^n\in(n)$, the ideal generated by $n$. But I don't see how this helps me.2012-06-27
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    @user30382: Dear user, Your first four lines are on the right track, but you then move in a direction that is not relevant. (Why would you want to solve for $x^n$?) Remember that you are trying to prove that $x = 0$ (so wouldn't it make more sense to solve for $x$?), and what you know is that some power of $x$ equals $0$. Take the equation $x^n + \cdots + n x = 0$, and use it to deduce that $x = 0$. (Why don't you begin by imagining that $x^2 = 0$, and see what you can say. After that, suppose that $x^3 = 0$ and see what happens. You should then see how to proceed in general.) Regards,2012-06-27
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    I see now, thank you for your help (and patience). I failed to notice that in the equation $$\sum_{i=1}^n{n\choose i}x^i=nx\cdot\sum_{i=0}^n{n\choose i+1}\frac{x^i}{n}=0,$$ the right most summation is of the form $1+y$ with $y\in\sqrt{0_A}$, and hence a unit.2012-06-28
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    @user30382: Dear user, You're welcome (and the observation about obtaining a unit is a nice way to see it; I had in mind a somewhat clunkier argument by induction on the degree of nilpotence of $x$). Best wishes,2012-06-28
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    @user30382: Dear user, One more thing; note that the assumption that $A$ is an order is crucial, because this allows you to deduce that $x=0$ from the *a priori* weaker statement that $n x =0$. (So really what is used is that $A$ is torsion-free as an abelian group.) Best wishes,2012-06-28
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    @user30382: P.S. Also, you have to be a little careful in how your phrase your argument, because $n$ does not dividie $n$ choose $i$ in general if $n$ is not prime (e.g. $4$ does not divide $6$, which is $4$ choose $2$). I have added a sketch of the argument to my answer, which phrases things slightly differently to avoid this minor annoyance.2012-06-28
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    Keen observation! I overlooked this minor detail myself. Your phrasing makes for a much nicer proof. And I never noticed that the sum of a zero divisor and a nilpotent is again a zero divisor.2012-07-01