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$\begingroup$

Let me refer you to:

http://www-users.math.umd.edu/~karpuk/chap3solns.pdf

Page 2, ex. 4

Can you please explain the following step:

tb=f(s')b=s'b

Why f(s')=s' ?

Thanks

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    @Zev Chonoles: sorry, just edited it.2011-04-29

1 Answers 1

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Hint: We have a map $f:A\rightarrow B$. This makes $B$ into an $A$-module. What is the definition of "$ab$", for $a\in A$ and $b\in B$?

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    @user 6495: Exactly right. Don't worry, happens to all of us :)2011-04-29