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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?

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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.)

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  • $\begingroup$ 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. $\endgroup$ – Matt Samuel Nov 3 '18 at 20:47
  • $\begingroup$ @Eric Wofsey, thank you for this, perhaps you can assist me with this further question: math.stackexchange.com/questions/2983549/… $\endgroup$ – Mariah Nov 4 '18 at 0:31

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