Proof of Hilbert's Nullstellensatz, weak form.
This proof is not sound as written, because certain polynomials (namely nonzero 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.
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Updated on July 23, 2020Comments

Admin over 2 years
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 ntuple 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!

Admin over 9 years@MattE Thanks for your answer! I suppose my assertions are correct for polynomials that do not become nonzero constants on equating all variables. As for the second point, in retrospect, all polynomials $f_1,f_2,\dots,f_n$ should have at least one variable $x_i$ in common in order to generate the unit ideal.

Ragib Zaman over 9 years@AyushKhaitan Even if you took $f_1=x_1, f_2=x_1x_2$ you don't get the unit ideal (evaluate at 0). It seems you are trying to use Bezout's identity, which holds for PIDs but not (much) more generally.

Admin over 9 yearsOh sorry I had meant the polynomials having an $x_i$ in common which do not have a common factor, and one is not a multiple of the other. These polynomials can be treated as polynomials in $x_i$ with coefficients in $R[x_1,x_2,\dots,x_{i1},x_{i+1},\dots,x_n]$. These polynomials I suppose should generate the unit by the Euclidean algorithm.

Matt E over 9 years@AyushKhaitan: Dear Ayush, How about $x_1  x_2$ and $x_1 + x_2$. These have the variables $x_1$ and $x_2$ in common, and are both irreducible with no common factor. They don't generate the unit ideal. Part of the difficulty with the Nullstellensatz is exactly that the algebra of the situation is not as simple as you are trying to make it seem. Maybe you should try thinking in a more geometric way (e.g. my example gives two lines crossing in a single point). There are sketches of various geometric proofs of the Nullstellensatz on MO, among other places. Regards,