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With AC, it is easy to see that any vector space is injective, and free, therefore alse flat and projective. Without AC, vector spaces can be not free. Are they must be projective modules? Flat modules? What about injectiveness?

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I gave a partial answer for projectivity. Unfortunately, I don't have a tight grasp on the notion of flatness, so I can't say whether or not the example I gave is flat or not. –  Asaf Karagila Apr 21 at 7:41
    
(Moreover, Wikipedia gives a long list of equivalents for flatness, but I'm fairly sure that the proof of some of these equivalence would require choice, so I'm not even clear what is the correct definition for flat in the absence of choice.) –  Asaf Karagila Apr 21 at 7:47

2 Answers 2

I think vector spaces must still be flat (tensor product is exact). I don't think any of the steps in the following proof use choice, although it's quite possible I'm mistaken:

Finite dimensional vector spaces are flat.

Every vector space is the filtered colimit of its finite dimensional subspaces.

Tensor products commute with colimits.

Filtered colimits of exact sequences of vector spaces are exact.

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How do you define flatness, then? –  Asaf Karagila Apr 21 at 14:43
    
$F$ is flat if $F\otimes -$ is an exact functor. –  Jeremy Rickard Apr 21 at 15:22
    
I suppose that my next question, if so, is how do you define "exact functor", but I'll look it up first. Thanks. –  Asaf Karagila Apr 21 at 15:23
    
Yeah, glossing over the definitions I found in Wikipedia, and what little memory I can utilize from algebra related courses, I think that you're right here about flatness. –  Asaf Karagila Apr 21 at 15:34

(I completely revamped my answer, the previous version (link) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.)

The answer is no. For projectivity and injectivity. The reason is that the axiom of choice is equivalent to the following statement:

If $V$ is a vector space, and $K$ is a subspace of $V$, then there exists a a direct complement for $K$, that is a subspace $S$ such that $S\cap K=\{0\}$ and $S\cup K$ is a generating set for $V$.

If every vector space is projective, consider the case that $V$ is a vector space, $K$ is a subspace and $T\colon V\to V/K$ is the quotient map. Then there is some $h\colon V/K\to V$ such that $h\circ T$ is the identity function. In particular, this means that $K+\operatorname{im}(h)=V$, but $h$ cannot map any nonzero vector into $K$ so this is a direct sum indeed.

Therefore the statement "Every vector space is projective" implies the axiom of choice.

Similarly for injectivity, take the identity function $K\to V$, then there is a projection map whose kernel is a direct complement for $K$.


I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.

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And of course this also answers the "injectivity" question. In fact, it shows that even finite-dimensional vector spaces need not be injective. –  Jeremy Rickard Apr 21 at 8:04
    
Ah, I see. Thanks for pointing that out! –  Asaf Karagila Apr 21 at 8:07

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