# Is $\{ 0 \}$ a basis of the free module $\{ 0 \}$?

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:

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.

• No, $\{0\}$ is not a free module on one generator. – Lord Shark the Unknown May 23 '19 at 17:13
• Can you explain how I can get to that conclusion using the definition I posted? – hidenoris May 29 '19 at 15:34

The set $$\{0\}$$ is not linearly independent, because $$r0=0$$ for any nonzero $$r\neq 0$$.
• It looks like $\{0\}$ is a basis over the null ring, though. – hmakholm left over Monica May 23 '19 at 17:29
• 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
• 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 1:56