Construction of tensor product of algebras $A \otimes_k B$ in the book "Algebraic geometry" by Milne

Question

$\underline{ \text{Tensor product of algebras}}:$ (J.S. Milne book Algebraic geometry, page $31$)

Let $A, B$ be two $k$-algebras. Now, J.S. Milne given the following construction of $A \otimes_k B$ in his book "Algebraic geometry".

Construction:

At first regard or assume $A,B$ as $k$-vector spaces and form the tensor product $A \otimes_k B$. There is a multiplication map $(A \otimes_k B) \times (A \otimes_k B) \to A \otimes_k B$ for which $$ (a \otimes b)(a' \otimes b')=aa' \otimes bb'.$$ This multiplication map makes $ A \otimes_k B$ into a ring and I understood this.

Next, he says that the homomorphism $$ c \mapsto c(1 \otimes 1)=c \otimes 1=1 \otimes c$$ makes $A \otimes_k B$ into a $k$-algebra. But how ? Please explain it. Because to be a algebra we need a vector space and additionally multiplication structure.

Next, he claims the maps $$ a \mapsto a \otimes 1: \ A \to C \ \text{and} \ b \mapsto 1 \otimes b: \ B \to C$$ are homomorphisms and these homomorphisms makes $A \otimes_k B$ into the tensor product of $A$ and $B$. But how? Please explain the last two paragraphs because I want to understand tensor product of algebras.

Answer 1

As it is constructed, $A \otimes_k B$ is a vector space with a binary operation. In order to show that $A \otimes_k B$ qualifies as an algebra, we need to show that it satisfies right and left distributivity and compatibility of the binary operation with scalar multiplication.

For left distributivity: note that an arbitrary element can be written as a sum of the form $\sum_{i=1}^n a_i \otimes b_i$ for some choice of $a_i \in A$ and $b_i \in B$. So, consider arbitrary elements $x,y,z \in A \otimes_k B$. We need to show that $$ z \cdot (x +y ) = z \cdot x + z \cdot y. $$ To show that this is the case, express each of $x,y,z$ as sums of simple tensors. That is, say $$ x = \sum_{i=1}^p a_i^x \otimes b_i^x, \quad y = \sum_{j=1}^q a_j^y \otimes b_j^y, \quad z = \sum_{k=1}^r a_k^z \otimes b_k^z. $$ The left hand side can be written as $$ z \cdot (x +y) = \left( \sum_{k=1}^r a_k^z \otimes b_k^z \right) \left( \sum_{i=1}^p a_i^x \otimes b_i^x + \sum_{j=1}^q a_j^y \otimes b_j^y \right). $$ By expanding all products, show that this coincides with the right side $$ z \cdot x + z \cdot y = \left( \sum_{k=1}^r a_k^z \otimes b_k^z \right) \left( \sum_{i=1}^p a_i^x \otimes b_i^x \right) + \left( \sum_{k=1}^r a_k^z \otimes b_k^z \right) \left( \sum_{j=1}^q a_j^y \otimes b_j^y \right). $$ That right-distributivity and compatibility with scalars hold can be shown in a similar fashion.

Answer 2

If $A,B$ are $k$-algebras spaces then $A,B$ are just rings with a $k$-vector space structure, namely you have maps $m_A:A\times k\longrightarrow A,m_B:B\times k\longrightarrow B$ that realise " scalar multiplication".

Then to put a $k$-vector space structure on $A\otimes_k B$ just define $(A\otimes _k B)\times k\longrightarrow A\otimes _k B$ by sending each $(a\otimes b, \lambda)$ to $m_A(a,\lambda)\otimes b$ which is the same as $a\otimes m_B(b,\lambda)$ by definition of tensor product (as quotient of the free vector space).

Regarding the last part of your question, Milne is checking that the object $A\otimes_k B$ satisfies the universal property of coproduct in the category of algebras:

Given $X,Y$ two objects in a category $\mathscr{C}$, an object $\Sigma$ is called coproduct of $X,Y$ if there exist morphisms $i_X : X \to \Sigma, i_Y : Y\to \Sigma$ such that the following universal property is satisfied: for all objects $Z$ and morphisms $j_X : X\to Z,,j_Y\to Z$ there exist an unique morphism $\varphi : \Sigma \to Z$ such that $j_X = \varphi : i_X, j_Y = \varphi \circ i_Y$.

Milne says that you can check this property (for $X=A,Y=B$ in $\mathscr{C}$ the category of commutative $k$-algebras) by defining $i_A$ and $i_B$ as the maps $a\mapsto a\otimes 1,b\mapsto 1\otimes b$ respectively.