Let R be a ring: since R is an abelian group under +, na has meaning for us for n in Z, a in R. Show that (na)(mb)=(nm)(ab) if n,m are integers and a,b in R.

Please show steps. Thanks!

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- Apr 24th 2009, 11:24 AMmpryalquotient Rings 1
Let R be a ring: since R is an abelian group under +, na has meaning for us for n in Z, a in R. Show that (na)(mb)=(nm)(ab) if n,m are integers and a,b in R.

Please show steps. Thanks! - Apr 24th 2009, 11:33 AMJhevon
you've posted a lot of questions, i can't help but think you want us to do your homework for you :p

anyway, use what na and mb means...

$\displaystyle (na)(mb) = \underbrace{(a + a + \cdots + a)}_{n \text{ times}} \underbrace{(b + b + \cdots + b)}_{m \text{ times}} = \cdots$

what does this have to do with quotient rings?