Alternating multilinear function satisfies $f(A)=\det(A)f(Id)$

Question

I've just seen a proof of the statement: "Given $\alpha$ in a commutative ring $K$ there is a unique alternating multilinear function $f$ with $f(Id)=\alpha$."

The determinant is defined as the unique $f$ such that $f(Id)=1$. I don't understand why for each alternating multilinear function $f$ we have $$f(A)=\det(A)f(Id)$$

I would appreciate if anyone could explain me why this is true. Thanks in advance.

Answer 1

The function $g$ defined by $g(A) = \det(A)f(\mathrm{Id})$ is alternating and multilinear. Since $$g(\mathrm{Id}) = \det(\mathrm{Id}) f(\mathrm{Id}) = f(\mathrm{Id}),$$ the statement you quoted implies that $g = f$.