show ring isomorphism

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April 4th 2010, 03:08 PM
dori1123

show ring isomorphism

Let $R$ be a ring and $M$ an $R$ -module.
$Hom_R(R,M)$ is the set of $R$ -homomorphisms from $R$ to $M$ .
Define $\phi: Hom_R(R,M) \rightarrow M$ by $\phi(f)=f(1)$ . Show $\phi$ is a ring isomorphism.
I know how to show it is a homomorphism and one-to-one, but having trouble with onto. Can I get some help?
April 4th 2010, 05:39 PM
aliceinwonderland

Quote:

Originally Posted by dori1123

Let $R$ be a ring and $M$ an $R$ -module.
$Hom_R(R,M)$ is the set of $R$ -homomorphisms from $R$ to $M$ .
Define $\phi: Hom_R(R,M) \rightarrow M$ by $\phi(f)=f(1)$ . Show $\phi$ is a ring isomorphism (R-module isomorphism ?).
I know how to show it is a homomorphism and one-to-one, but having trouble with onto. Can I get some help?

Your M seems to be a unitary R-module from your question because f(1) is defined.

I think it is difficult to deduce that $\phi$ is onto by using a $\phi$ alone.

Rather, define a map $\psi:M \rightarrow Hom_R(R,M)$ given by $m \mapsto f_m$ , where $f_m(r) = rm$ .

Suppose $m_1 = m_2$ ( $m_1, m_2 \in M)$ . Then, $\psi(m_1) = f_{m_{1}}=f_{m_{2}}=\psi(m_2)$ , because $f_{m_{1}}(r)=rm_1=rm_2=f_{m_{2}}(r)$ . Thus $\psi$ is well-defined. I'll leave it to you to show that $\psi$ is an R-module homomorphism.

Then $\phi\psi=1_M$ and $\psi\phi=1_{Hom_{R}(R,M)}$ (verify this).
April 4th 2010, 07:24 PM
FancyMouse

Just to show that given any element x in M, you can construct a well-defined R-homomorphism f from R to M such that f(1)=x. The $f_m$ function that aliceinwonderland constructed is exactly demonstrating this idea.