How to compute the rank of the tensor product of two abelian groups?

Question

I tried to find a formula on the internet for the rank of the tensor product of two abelian groups -- e.g. $rank(\mathbb{Z}\otimes \mathbb{Z})$. If there's one, could you share it with me (with some details before if you want) before I attempt a calculation/proof?

Thank you in advance!

Answer 1

Let's first assume you are fine with the following properties of tensor products of abelian groups:

• $\mathbb{Z} \otimes A \cong A$ for any abelian group $A$
• $A \otimes B \cong B \otimes A$ for any abelian groups $A$ and $B$
• $\mathbb{Z}_m \otimes \mathbb{Z}_n \cong \mathbb{Z}_{\gcd(m,n)}$
• Distributivity of $\otimes$ over $\oplus$ (up to iso, of course).

Then, for any f.g. abelian groups $A$ and $B$, you can use the fundamental theorem of such groups to compute their tensor products.

Suppose $A$ has rank $r$ and $B$ has rank $s$. We can write $A \cong F_A \oplus \mathbb{Z}^r$ and $B \cong F_B \oplus \mathbb{Z}^s$ for finite abelian groups $F_A$ and $F_B$ (which are all finite direct sums of $\mathbb{Z}_k$'s for various $k$). You can then use the properties above to show that $$ \mathrm{rank}(A \otimes B) = rs = \mathrm{rank}(A) \cdot \mathrm{rank}(B) $$ just by distributing everything. The point is that the finite stuff will never contribute to the rank by the first and third bulleted properties above.