Short Exact Sequences Ending in Free Modules Split
Point of Post: We show, in later language, that every free module is projective. Or, in different terms, every short exact sequence sending in a free module splits.
We have seen that every short exact sequence of vector spaces splits and since free modules are “like” vector spaces, we should be able to make a similar statement for them. Indeed, we go one step further (which we could have done for vector spaces, but didn’t). We shall show that whenever we have a short exact sequence of the form
whre is free, then the sequence splits. The proof shall be very similar to the case for vector spaces, basically being a consequence of defining an -map on a basis.
Short Exact Sequence Ending in Free Modules Split
So, let’s just get right down to the theorem at hand:
Theorem: Let be a ring and
a short exact sequence of -modules. Then, this chain splits.
Proof: By the splitting lemma it suffices to show that there is a backmap for . To do this, fis a basis for . Define a map by defining to be any element which maps to and extending by linearity. Clearly this is an -map and we note that for any one has that
and so , and so is the desired backmap for . The conclusion follows.
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