Let \(k\) be a field.
State Hilbert’s Nullstellensatz Theorem.
Prove that if \(I=\langle f_1,\ldots,f_s\rangle\subseteq k[x_1,\ldots,x_n]\) is an ideal, then \(f\in\sqrt{I}\) if and only if \(1\in\langle f_1,\ldots,f_s,\;1-yf\rangle\subseteq k[x_1,\ldots,x_n,y]\).
Prove that if \(f\in k[x_1,\ldots,x_n]\) and \(J=\langle f\rangle\) is the principal ideal generated by \(f\), then \(\sqrt{J}=\langle f_1f_2\cdots f_r\rangle\) where \(f=f_1^{a_1}f_2^{a_2}\cdots f_r^{a_r}\) is the factorization of \(f\) into a product of distinct irreducible polynomials in \(k[x_1,\ldots,x_n]\).
Solution
1. Hilbert’s Nullstellensatz. If \(k\) is algebraically closed and \(I\subseteq k[x_1,\ldots,x_n]\) is an ideal, then\[ I(V(I))=\sqrt I. \]Equivalently, a polynomial \(f\) vanishes on every common zero of \(I\) in \(k^n\) if and only if \(f^N\in I\) for some \(N\ge1\).The weak form says that every maximal ideal of \(k[x_1,\ldots,x_n]\) is\[ (x_1-a_1,\ldots,x_n-a_n) \]for a unique \((a_1,\ldots,a_n)\in k^n\).
This is the standard strong, respectively weak, form of Hilbert’s Nullstellensatz. The algebraic-closedness hypothesis is essential for these formulations over \(k^n\).
Put \[ R=k[x_1,\ldots,x_n], \qquad K=\langle f_1,\ldots,f_s,1-yf\rangle\subseteq R[y]. \]
2. If \(f\in\sqrt I\), then \(1\in K\).
Choose \(N\ge1\) such that \(f^N\in I=(f_1,\ldots,f_s)\). Then \[ 1-(yf)^N=(1-yf)\left(1+yf+\cdots+(yf)^{N-1}\right). \] Hence \[ 1=(1-yf)\left(1+yf+\cdots+(yf)^{N-1}\right)+y^Nf^N. \] The first term lies in \((1-yf)\) and the second lies in \(IR[y]\), because \(f^N\in I\). Thus \(1\in K\).
3. If \(1\in K\), then \(f\in\sqrt I\).
If \(f=0\), then \(f\in\sqrt I\) trivially. Assume \(f\ne0\). Localize \(R\) at the powers of \(f\). There is an \(R\)-algebra homomorphism \[ \varphi:R[y]\longrightarrow R_f, \qquad \varphi(y)=f^{-1}. \] It sends \[ 1-yf\longmapsto 0. \] If \(1\in K\), applying \(\varphi\) therefore gives \[ 1\in IR_f. \] Thus there are elements \(a_i/f^{m_i}\in R_f\) such that \[ 1=\sum_i \frac{a_i}{f^{m_i}}f_i. \] Multiplying by a sufficiently large power \(f^N\) clears denominators and yields \[ f^N=\sum_i b_i f_i\in I. \] Hence \(f\in\sqrt I\).
4. Therefore\[ \boxed{ f\in\sqrt I \iff 1\in\langle f_1,\ldots,f_s,1-yf\rangle\subseteq R[y].} \]
3. This proves part (b).
For part (c), write \[ f=u f_1^{a_1}\cdots f_r^{a_r} \] with \(u\in k^\times\), the \(f_i\) pairwise nonassociate irreducibles, and \(a_i\ge1\). The unit \(u\) does not affect the principal ideal. Put \[ g=f_1\cdots f_r. \]
5. One has\[ (g)\subseteq\sqrt{(f)}. \]
Let \[ N=\max\{a_1,\ldots,a_r\}. \] Then \[ g^N=f_1^N\cdots f_r^N \] is divisible by \(f_1^{a_1}\cdots f_r^{a_r}\), hence \(g^N\in(f)\). Therefore \(g\in\sqrt{(f)}\). Since the radical is an ideal, \((g)\subseteq\sqrt{(f)}\).
6. Conversely,\[ \sqrt{(f)}\subseteq(g). \]
Let \(h\in\sqrt{(f)}\). Then \(h^N\in(f)\) for some \(N\), so \[ f_1^{a_1}\cdots f_r^{a_r}\mid h^N. \] The polynomial ring \(k[x_1,\ldots,x_n]\) is a UFD, so every irreducible \(f_i\) is prime. Thus \(f_i\mid h^N\) implies \(f_i\mid h\). Because the \(f_i\) are pairwise nonassociate primes, their product divides \(h\): \[ f_1\cdots f_r\mid h. \] Hence \(h\in(g)\).
7. Therefore\[ \boxed{\sqrt{(f)}=(f_1f_2\cdots f_r)}. \]
6. This proves part (c).