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I basically have to show that $d(R/(e))\cong R/(\frac{e}{g})$ where $R$ is a PID, and $\gcd(e,d)=(g)$.

I defined $\pi:d(R/(e))\rightarrow R/(\frac{e}{g})$ by $\pi:dr+dce\mapsto \frac{dr+dce}{g}$. Note that even though we are not in a field, we can divide this elements by $g$ since it divides $d$ and $e$. The function is clearly a homomorphism so I need to show that it is injective and surjective.

For injectivity I basically say that if $$\frac{d}{g}r+dc\frac{e}{g}\in \left(\frac{e}{g}\right)$$ happens if $$\frac{d}{g}r\in \left(\frac{e}{g}\right)$$ This is equivalent to saying that for some $k\in R$: $$\frac{d}{g}r=k\frac{e}{g}$$ Thus, $dr=ke\in (e)$, ergo, $dr+dce$ is zero in $d(R/(e))$ meaning that the kernel is trivial and thus the function is injective.

For surjectivity I let $r+c\frac{e}{g}\in R/(\frac{e}{g})$, then I want to find $r_1,c_1\in R$ such that $dr_1+dc_1e\mapsto r+c\frac{e}{g}$, but I get something like $$r_1=r\frac{g}{d}$$$$c_1=\frac{c}{d}$$The problem that I have is that dividing by $d$ in this case might not be in my $R$, so I am rather stuck here.

Any help?

Thanks

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Wait...what do you mean by $d(R/(e))$? Do you mean the ideal of $R/(e)$ generated by $d$? If so, then this is a ring without identity (unless $d\in U(R)$). –  user5137 Mar 8 '12 at 21:45
    
I believe it is $d$ multiples of $R/(e)$ and I should have mentioned this, we are viewing them as $R$ modules, not rings as their own. –  Daniel Montealegre Mar 8 '12 at 21:50
    
No, I am not making my self clear. Elements in $R/(e)$ times $d$. That is, $d(r+ce)$ –  Daniel Montealegre Mar 8 '12 at 21:54
    
Are you trying to show they are isomorphic as modules, or as rings? –  Arturo Magidin Mar 9 '12 at 4:17

1 Answer 1

up vote 4 down vote accepted

If you instead let $\pi:R\rightarrow d(R/(e))$ (as $R$-modules) by $\pi(r)=dr+(e)$, then $\pi$ is clearly an $R$-module homomorphism and is clearly onto.

So, suppose that $r\in ker(\pi)$. Then $dr+(e)=(e)$, and hence $dr\in (e)$, we have that $dr=\alpha e$ for some $\alpha\in R$. Keeping in mind that $\frac{e}{g},\frac{d}{g}\in R$, we have $\frac{d}{g}r=\alpha \frac{e}{g}$. And since $gcd\left(\frac{d}{g},\frac{e}{g}\right)=1$ and $\frac{d}{g}\vert \alpha \frac{e}{g}$ (if you aren't convinced of either of these things, sit down and write out a proof to convince yourself), it follows that $\frac{d}{g} \vert \alpha$, hence $r\in \left(\frac{e}{g}\right)$ and $ker(\pi)\subseteq \left(\frac{e}{g}\right)$. I'll leave it to you to show that $\left(\frac{e}{g}\right)\subseteq ker(\pi)$.

Thus, by the First Isomorphism Theorem, $R/(e/g)\cong d(R/(e))$.

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Awesome! That is much easier. Thanks :) –  Daniel Montealegre Mar 8 '12 at 22:20

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