Dummit and Foote Page :$544$ Proposition $30$

I have a little bit confusion in Dummit and Foote Book on the Page No $$:544$$

Proposition $$30$$: For any field $$F$$ there exist an algebraically closed field $$K$$ containing $$F$$

Proof :

My confusion : It is given that $$g_(x_1,x_2,..,x_n,x_{n+1}....,x_m)f_1(x_1) +.......+g_n(x_1,x_2,..,x_n,x_{n+1}...,x_m)f_n(x_n)=1$$

Now if $$x_{n+1}=....=x_m=0$$ then $$g_(x_1,x_2,..,x_n,0....,0)f_1(a_1) +.......+g_n(x_1,x_2,..,0,0...,0)f_n(a_n)=0$$

Here im not getting that why $$g_n(x_1,x_2,..,x_n,0....,0)=0?$$

• It is the $f_k(\alpha_k)$ the ones that are zero, for $k=1,2,...,n$.
– plop
Commented Jun 23, 2021 at 22:17
• Please don't rely on pictures of text. Commented Jun 23, 2021 at 22:33
• Im sorry for that @Shaun Actually ,it is very Long Proof.So i decided to make screenshot Commented Jun 23, 2021 at 22:41

The proof is not claiming that $$g_i(x_1,\ldots,x_n,0,\ldots,0)=0$$.
Its point is that $$f_i(x_i)$$ is $$0$$, because the $$x_i=\alpha_i$$s were chosen to make this true.
So no matter what the value of $$g_i(\cdots)$$ is, it will be killed when you multiply it by $$f_i(x_i)$$.
• thanks you @Troposphere for saving my time .Actually I misunderstood and I have wasted more than $1$ hours thinking about why $g_n(x_1,...x_n,0.,0..)=0?$ Commented Jun 23, 2021 at 22:26