4

Let $A,B$ be $k$-algebras and $A \otimes_k B$ be their tensor product (over $k$). I want to show that $(a \otimes b)(a' \otimes b')=(aa' \otimes bb')$ is distributive (because I need to show that $A \otimes_k B$ is a ring with this multiplication). So the aim is to get $(a \otimes b)((a' \otimes b')+(a'' \otimes b''))=((a \otimes b)(a' \otimes b'))+((a \otimes b)(a'' \otimes b''))$. Now $(a \otimes b)(a' \otimes b')+(a \otimes b)(a'' \otimes b'')=$ after fiddling around $(a \otimes b)((a+a') \otimes (b+b'))-(a \otimes b)(a' \otimes b'')-(a \otimes b)(a'' \otimes b'')=?$ I am stuck, how can I proceed?

Lucia
  • 41
  • 1

3 Answers3

3

How are you defining the product?

You've only specified the value of the product on pure tensors, but usually, "most" of the elements of $A \otimes_k B$ cannot be written in the form $a \otimes b$. Also, you haven't checked that the definition you give is well-defined.

Usually, products like that are defined in this manner as "the distributive operation that has this value on the pure tensors", so your definition is somewhat begging the question if you don't know such a thing exists.

My suggestion is to think about what you know about bilinear functions on modules. In the way I would go about it, the module $(A \otimes_k B) \otimes_k (A \otimes_k B)$ plays a major role, since your (alleged) product is equivalent to a homomorphism from this to $A \otimes_k B$.

0

I have checked the definition and it is well defined, (on pure tensors). I also know that if $X$ is a basis for $A$ and $Y$ is a basis for $B$, then $\{x \otimes y: x \in X, y\in Y \}$ is a basis for the tensor product. This definition is what I was given and I have seen it in many textbooks. However, in all these textbooks, they claim that this operation is OBVIOUSLY distributive.

Lucia
  • 41
  • 1
  • Don't use bases to the definition unless you are a physicist. :) Seriously, the distributivity should be regarded as a special aspect of bilinearity: construct a $k$-bilinear mapping $\mu \colon (A \otimes_k B) \times (A \otimes_k B) \rightarrow A \otimes_k B$ such that $\mu(a \otimes b, a' \otimes b') = aa' \otimes bb'$. Do you know how to take the tensor product of two linear maps? – KCd Nov 30 '13 at 06:55
0

One easy way to do that is to use the The universal property of tensor product:

Define a map $\varphi:A\times B \times A \times B \to A\bigotimes B$ by:

$\varphi(a.b,a',b')=aa'\bigotimes bb'$ is multilinear over $k$.

It easy to check, for instance:

$\begin{align*} \varphi(a,r_1b_1+r_2b_2,a',b') &=aa'\bigotimes \left ( r_1b_1+r_2b_2 \right )b' \\ &=aa'\bigotimes r_1b_1b' +aa'\bigotimes r_2b_2b' \\ &= r_1\varphi(a,b_1,a',b')+r_2\varphi(a,b_2,a',b') \end{align*}$

So $\varphi$ induces a $k-module$ homomorphism $\Phi$ from $A\bigotimes B\bigotimes A \bigotimes B$ to $A\bigotimes B$ with:$\Phi(a\bigotimes b\bigotimes a' \bigotimes b')=aa'\bigotimes bb'$

Next, thinking $A\bigotimes B\bigotimes A \bigotimes B$ as $(A\bigotimes B)\bigotimes (A \bigotimes B)$.

Now, repeat the above process to attain a well-defined $k$-bilinear map $\psi$ from $(A\bigotimes B)\times (A \bigotimes B)$ to $A\bigotimes B$ with $\psi(a\bigotimes b,a'\bigotimes b')=aa'\bigotimes bb'$ which implies that our multiplication is indeed well-defined.

An Khuong Doan
  • 2,107