# Additive group of tensor product of commutative $k$-algebras with units

Let $$A_1, A_2$$ denote two unital, commutative $$k$$ algebras, where $$k$$ is a field. Then by definition $$A_i$$ are rings with a bilinear operation on $$k$$.

When we speak of the tensor product $$A_1 \otimes A_2$$, this is also a unital commutative ring.

I want to make sure I understand correctly; is the additive abelian group of $$A_1 \otimes A_2$$ just $$\{ 0\}$$? How do we define addition on this ring?

If $$a_1 \otimes a_2, b_1 \otimes b_2 \in A_1 \otimes A_2$$ and their addition would defined as $$a_1 \otimes a_2 + b_1 \otimes b_2 = a_1 + b_1 \otimes a_2 + b_2$$, then $$a_1 \otimes a_2 = a_1 \otimes 0 + 0 \times a_2 = 0$$, no?

How do we define addition in this ring, and do I misunderstand something more basic?

No, $$A_1\otimes A_2$$ is not just $$\{0\}$$ (unless either $$A_1$$ or $$A_2$$ is just $$\{0\}$$). It is not correct that $$a_1 \otimes a_2 + b_1 \otimes b_2 = (a_1 + b_1) \otimes (a_2 + b_2)$$. In general, there is no way to simplify an expression like $$a_1 \otimes a_2 + b_1 \otimes b_2$$ in the tensor product. Keep in mind that a general element of the tensor product is not of the form $$a\otimes b$$; instead it is a sum of elements of this form.
To understand the additive structure of $$A_1\otimes A_2$$, the following theorem is very useful. If $$B_1$$ is a basis for $$A_1$$ and $$B_2$$ is a basis for $$A_2$$, then the set of elements of $$A_1\otimes A_2$$ of the form $$e\otimes f$$ for $$e\in B_1$$ and $$f\in B_2$$ is a basis for $$A_1\otimes A_2$$. (This theorem holds more generally for tensor products of free modules over any commutative ring.)
• I think pointing out that $(a_1+a_2)\otimes b_1=a_1\otimes b_1+a_2\otimes b_1$ etc. would be helpful since the OP has no idea about the structure of the additive group. Nov 3, 2018 at 20:47