# ring morphism and modules

• Nov 7th 2008, 09:13 PM
roporte
ring morphism and modules
Let A and B rings, M a B-module and $\displaystyle \phi: A \rightarrow B$ a ring morphism. Prove that the action $\displaystyle a._{\phi}x=\phi(a)x$ defines a structure of A-module over M.

thanks!!
• Nov 8th 2008, 12:42 AM
NonCommAlg
Quote:

Originally Posted by roporte
Let A and B rings, M a B-module and $\displaystyle \phi: A \rightarrow B$ a ring morphism. Prove that the action $\displaystyle a._{\phi}x=\phi(a)x$ defines a structure of A-module over M.

thanks!!

the action is clearly well-defined because if a = a' and x = x', then $\displaystyle \phi(a)x=\phi(a')x=\phi(a')x'.$ now M is a B-module and $\displaystyle \phi$ is a homomorphism. therefore:

$\displaystyle (a + b)\cdot x = \phi(a+b)x=(\phi(a) + \phi(b))x=\phi(a)x + \phi(b)x=a \cdot x + b \cdot x.$ similarly: $\displaystyle a \cdot (x+y)=\phi(a)(x+y)=\phi(a)x + \phi(a)y=a \cdot x + a \cdot y.$ finally:

$\displaystyle (ab) \cdot x = \phi(ab)x=(\phi(a)\phi(b))x=\phi(a)(\phi(b)x)=a \cdot (b \cdot x). \ \ \Box$