Deduce the Bezout bound from Bernstein's Theorem: Show that if polynomials $f_1,f_2,\dots,f_n \in \mathbb{C}[x_1,\dots,x_n]$ have finitely many solutions in $(\mathbb{C}^*)^n$, then the number of solutions is at most $\deg(f_1)\cdot\deg(f_2)\cdots\deg(f_n)$.

### Solution:

Let the polynomial $f_i$ in $n$ variables have degree $d_i$. Then the largest possible polytope for $New(f_i)$ is the n-dimensional simplex with side length $d_i$, which we will refer to as a $(d_i,n)$-simplex. This corresponds to having every monomial of degree $d_i$ in the support of $f_i$. So we know $New(f_i)$ is contained within this simplex.

We now recall some facts about simplices. First of all, the Minkowski sum of two n-simplexes, with side lengths a and b respectively, is the n-simplex with side length $a+b$. Second, the formula for the volume of a $(c,n)$-simplex is $1/n!(c)^n$. This can be seen explicitly by integrating.

Now we are ready to prove the Bezout bound. By Bernstein's Theorem, we know that given these hypotheses, the number of solutions is at most the mixed volume of $New(f_1),..,New(f_n)$. To calculate this, we want the coefficient of $\Pi \lambda_i$ in $Vol(\sum \lambda_iP_i)$.

We see $Vol(\sum \lambda_iP_i) \leq Vol(\sum \lambda_i ((d_i,n)-simplices))=Vol(\sum (\lambda_id_i,n)-simplices)=Vol((\sum \lambda_id_i,n)-simplex)$, using first the fact that each Newton polytope is contained in some simplex, and then the facts about simplices given above. We now just need the volume of a simplex, which is $1/n!(\sum(\lambda_id_i))^n$. Since all we need is the coefficient of $\Pi \lambda_i$, we use a counting trick to see that this coefficient in $(\sum(\lambda_id_i))^n$ is $n!\Pi d_i$. (This is not really a trick, same method used in exercise 3, you can just think about writing out the sum n times, and counting all the ways to get all the lambda's. There are n choices from the first sum, n-1 from the second, etc. The product of $d_i$'s comes from the fact that it was a sum of $\lambda_id_i$'s.) So we see that the mixed volume of a collection of $(d_i,n)$-simplices is $\Pi d_i$, the product of degrees of all the polynomials. Following our string of inequalities, we now see that the number of zeroes is bounded by this product.