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Let $F$ a field and $V$ an $F$-module (i.e. an $F$-vector space). Then $V$ has a basis, and so $V$ is free. All free modules are projective, so $V$ is projective too.

Is this ok?

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  • $\begingroup$ Yes, it is OK. The main point of course is this question. $\endgroup$ – Dietrich Burde Dec 17 '17 at 19:52
  • $\begingroup$ "ok proof" = "proof"... $\endgroup$ – Jean Marie Dec 17 '17 at 19:53
  • $\begingroup$ @JeanMarie I of course meant "ok" as in "valid" $\endgroup$ – Alex Dec 17 '17 at 19:54
  • $\begingroup$ As a generalization, all modules over a division ring are also projective. $\endgroup$ – stressed out Dec 17 '17 at 20:33
  • $\begingroup$ Having all modules projective characterizes semisjmple rings. Having all modules free characterizes division rings. $\endgroup$ – rschwieb Dec 17 '17 at 22:37

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