I am currently studying for a final exam and am confused about the proof of $\det(AB)=\det(A)\det(B)$ given in Hoffman/Kunze. I'll type out the entire thing so that my question will be in the correct context.
Theorem: Let $K$ be a commutative ring with identity, and let $A$ and $B$ be $n \times n$ matrices over $K$. Then, $\det(AB)=\det(A)\det(B)$. Proof:
Let $B$ be a fixed $n \times n$ matrix over $K$, and for each $n \times n$ matrix $A$ define $D(A)=\det(AB)$. If we denote the rows of $A$ by $\alpha_{1},...,\alpha_{n}$, then $D(\alpha_{1},...,\alpha_{n})=\det(\alpha_{1}B,...,\alpha_{n}B)$. Here $\alpha_{j}B$ denotes the $1 \times n$ matrix which is the product of the $1 \times n$ matrix $\alpha_{j}$ and the $n \times n$ matrix $B$. Since, $(c \alpha_{i} + \alpha_{i}')B = c \alpha_{i}B + \alpha_{i}' B$ and $\det$ is $n$-linear, it is easy to see that $D$ is $n$-linear. If $\alpha_{i}=\alpha_{j}$, then $\alpha_{i}B=\alpha_{j}B$, and since $\det$ is alternating, $D(\alpha_{1},...,\alpha_{n})=0$. Hence $D$ is alternating. Now, $D$ is an alternating $n$-linear function, and so $D(a)=\det(A)D(I)$, but $D(I)=\det(IB)=\det(B)$ and thus, $\det(AB) = D(A) = \det(A)\det(B)$.
I think I understand the entire proof, how you show it and why each step is taken, but I am just confused about how the part right before "it is easy to see that $D$ is $n$-linear" show that $D$ is $n$-linear. If someone could just fill in the details or give an explanation of why that is sufficient would be helpful.
Thank you.