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These is an equivalent relation about projective modules. P is projective , (1)P is a direct summand of free module (2)If P is a quotient of the R-module M, then P is isomorphic to direct summand of M.

I am confused here, what does it mean that P is a quotient of the R-module M, the quotient module of M? Am I right with here?

user53800
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1 Answers1

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This means that if $\,P\cong M/N\,$ , then $\,P\,$ is isomorphic to a direct summand of o $\,M\,$ , or what is the same: if we have a short exact sequence

$$0\longrightarrow N\longrightarrow M\stackrel{\pi}\longrightarrow P\longrightarrow 0$$

then the sequence splits...which follows directly from the definition of projective, since then we can find a homom. $\,f:P\to M\,$ s.t. $\,\pi\circ f=Id_P\,$ .

DonAntonio
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  • Here the sequence spilits, that means $M=P\otimes N$, so M is a direct summand, but I can not see clearly how P is a direct summand – user53800 Jan 05 '13 at 00:38
  • No @user53800 , in $,M=P\times N,$ , it is $,P,$ which is a direct summand... – DonAntonio Jan 05 '13 at 00:40
  • oh I see, P is a direct summand as a component, is that right? I think I　ｔｈｉｎｋ　ｉｔ　ｉｎ　ａ　ｗｒｏｎｇ　ｗａｙ – user53800 Jan 05 '13 at 00:42
  • Indeed @user53800, though perhaps it is more comment the name of "factor" or "direct, free, whatever factor" – DonAntonio Jan 05 '13 at 00:43
  • Thanks so much, this has confused me for a long time, because I always get wrong result by using that, :) – user53800 Jan 05 '13 at 00:46
  • Anytime, @user53800 . You could always use some algebra book or the net as consultations sources, as indeed some of this stuff may be confusing. – DonAntonio Jan 05 '13 at 00:56