$\underline{ \text{Tensor product of algebras}}:$ (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.
Best Answer
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:
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.