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Here is what I have so far:

Consider $\mathbb{C}[x]/(x)$. This ring is all complex coefficient polynomials in $x$, and when quotient by the ideal $(x)$, we are subsituting $0$ for every $x$ in the polynomial, which kills all terms except the zero degree term, thus $\mathbb{C}[x]/(x) \cong \mathbb{C}$. Similarly, when we quotient by the ideal $(x-1)$ we set every $x$ in the polynomial to $1$, and we again have that $\mathbb{C}/(x-1) \cong \mathbb{C}$. Call the module $\mathbb{C}[x]/(x)=M$ and the other $N$. The difference between modules $M$ and $N$ is in the multiplication by the ring $\mathbb{C}[x]=R$. Let $p(x)\in R$ and let $m\in M$ and $n\in N$. $p(x)\cdot m=p(0)m$ while $p(x)\cdot n=p(1)n$. Thus even thought $M$ and $N$ are equivalent set wise they behave differently as modules over the same ring $R$. A homomorphism $\phi \in \operatorname{Hom}(M,N)$ must satisfy for all $m, m' \in M$ and $p, q \in R$, $\phi(p(x)m+q(x)m')=p(x)\phi(m)+q(x)\phi(m')$ Consider the $(1-x)\in R$, we know that $\phi$ is determined by where it send 1, since for any $c\in\mathbb{C}$, $\phi(c)=\phi(1c)=c\phi(1)$ since $\phi$ is homomorphism over $R$. We see that $\phi(1)=\phi((1-0)1)=\phi((1-x)1)=(1-x)\phi(1)=(1-1)\phi(1)=0\phi(1)=0$. Thus there is only the trivial homomorphism and $\operatorname{Hom}(M,N)$ is the zero module.

Is this correct?

Also,is it usually the case that for a given ring $R$ and two distinct non-unit elements $r, r' \in R$ the module of homomorphisms from the module $R/(r)$ to $R/(r')$ is isomorphic to $R/(\gcd(r,r'))$.

For example if $R=\mathbb{Z}$ and $M=R/(8)$ and $N=R/(12)$ then $\operatorname{Hom}(M,N)\cong R/(\gcd(8,12)) = R / (4)$.

Thanks!

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Hint:$$ \operatorname{Hom}(R/a,M)=\{m|am=0\} $$ –  Blah Apr 7 '12 at 20:25
    
@Blah, could you please expand. Are you using the same $R$ and $M$? –  Steven-Owen Apr 7 '12 at 20:45
    
No, this is quite general: $$\operatorname{Hom}(R/a,M)\rightarrow \{m|am=0\} \quad f \mapsto f(1)$$ is an isomorphism for any commutative ring $R$ and module $M$ –  Blah Apr 8 '12 at 8:12

1 Answer 1

Yes it looks correct. There is another way to think about it too: take a look at what are called comaximal ideals.

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