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I'm studying modules by reading Dummit and Foote, and I'm having a problem understanding the definition of a free module. I read this stackexchange question, but I couldn't figure it out.

The textbook defines a free module as following:

definition of a free module

The following is the example that I'm confused about.

Let $F = \{ 0 \}$. Then $F$ is a $\mathbb{Z}$-module since $F$ is an abelian group under addition.

In the stackexchange question, the empty set is given as the basis for $F$. That makes sense because $F$ has no nonzero elements, so it satisfies the definition vacuously.

But, using the same logic, wouldn't $F$ be a free $\mathbb{Z}$-module on the set $\{ 0 \}$?

I think that would be a problem because if that was the case, the rank of $\{ 0 \}$ would be both 0 and 1.

hidenoris
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1 Answers1

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The set $\{0\}$ is not linearly independent, because $r0=0$ for any nonzero $r\neq 0$.

The empty set does span the zero module, because an empty sum is zero.

Ehsaan
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  • It looks like ${0}$ is a basis over the null ring, though. – hmakholm left over Monica May 23 '19 at 17:29
  • I don't think the definition I posted (the one from Dummit and Foote) mentions anything about linearly independent sets. – hidenoris May 23 '19 at 18:16
  • That definition is equivalent to: (1) $A$ spans $F$, and (2) $A$ is linearly independent. Think about it: clearly $1\cdot 0 = 0 = 0\cdot 0$, so $0$ does not have a unique representation as a combination of elements of ${0}$. – Ehsaan May 23 '19 at 18:54
  • I'm not sure if that answers my question. The definition I posted states "nonzero element x of F, there exist..." – hidenoris May 23 '19 at 20:43
  • Okay, it's just a weird definition. I think most algebraists would agree that a basis is a linearly independent spanning set. A free module with a finite basis should be isomorphic to $R^n$, but this fails if you're using D&F's definition. Rank should be the cardinality of a basis, and should uniquely determine a free module (over a commutative ring). – Ehsaan May 24 '19 at 01:56
  • So, with D&F's definition, ${ 0 }$ is indeed a basis of the free module ${ 0 }$? – hidenoris May 29 '19 at 15:34
  • Sure. But be wary that now you can have non-isomorphic free modules of the same rank (namely ${0}$ and $R$ both have rank $1$). – Ehsaan May 30 '19 at 02:42