The statement of Hilbert's Nullstellensatz, weak form, as given here is "Let $f_1,f_2,\dots,f_n$ be polynomials in $K[x_1,x_2,\dots,x_n]$, where $K$ is an algebraically closed field. Then $1=\sum{g_t f_t}$ for suitable $g_t\in K[x_1,x_2,\dots,x_n]$ if and ony if the algebraic variety $V(f_1,f_2,\dots,f_n)\in \Bbb{K}^n =\emptyset$"
I'd like to start off by proving that if $K$ is an algebraically closed field, then any polynomial $f\in K[x_1,x_2,x_3,\dots x_n]$ has a solution in $\Bbb{K}^n$. We know that any polynomial in $K[x]$ has a solution in $K$. Let us take $f\in K[x_1,x_2,x_3,\dots x_n]$. If we have $x_1=x_2=\dots x_n$, then this becomes a polynomial in $K[x_1]$. This has a solution in $K$. Hence, the original $f$ has a solution in $\Bbb{K}^n$, where all the elements in the n-tuple are equal.
Proof of Hilbert's Weak Nullstellensatz:
If $V(f_1,f_2,\dots f_n)=\emptyset$, then not all the polynomials can have the same factor in common. If all polynomials have a factor in common, then that polynomial will have a solution in $\Bbb{K}^n$. This contradicts the fact that $V(f_1,f_2,\dots f_n)=\emptyset$. Hence, $1=\sum{g_t f_t}$.
If $1=\sum{g_t f_t}$, then all the polynomials do not have any factor in common. Hence, $V(f_1,f_2,\dots f_n)=\emptyset$.
Is this proof sound? The proof that I'm currently reading is vastly different from this.
Thanks in advance!
Best Answer
This proof is not sound as written, because certain polynomials (namely non-zero constant ones) don't have zeroes in $K$. E.g. if $n = 2$ and $f = 1 + x_1 - x_2$, and you set $x = x_1 = x_2$, this reduces to the constant polynomial $1$, which has no zero in $K$.
The second step is also incorrect. E.g. in $K[x_1,x_2],$ the polynomials $x_1$ and $x_2$ have no factor in common, but they don't generate the unit ideal.