This post is about Hilbert's basis theorem and the Nullstellensatz (theorem of zeros), which together form the foundational results of commutative algebra––the algebraic part of algebraic geometry––and serve to show that any set of polynomials not generating the unit ideal in $F[t_1, \dots , t_n]$, with $F$ an algebraically closed field, have a common zero.

Proposition 1. Let $R$ be a ring and $M$ an $R$-module. Then the following are equivalent: (1) Every submodule of $M$ is finitely generated.
(2) $M$ satisfies the ascending chain condition, i.e., if $M_i \subset M$ are submodules and $$M_1 \subset M_2 \subset \cdots \subset M_n ⊂ \cdots ,$$ then there exists a positive integer $N$ such that $M_N = M_{N+i}$ for all $i \geq 0$. We say every ascending chain of submodules of $M$ stabilizes. Equivalently, there exists no infinite chain
$$M_1 < M_2 < \dots < M_n < \cdots .$$ (3) $M$ satisfies the Maximum Principle, i.e., if $S$ is a non-empty set of submodules of $M$, then S contains a maximal element, that is a module $M_0 \in S$ such that if $M_0 ⊂ N$ with $N \in S$, then $N = M_0$.
An $R$-module $M$ satisfying any (all) of these equivalent conditions is called a Noetherian $R$-module or $R$-Noetherian.

Proof. $(1) \Rightarrow (2)$: Let $$\mathcal{C}: M_1 \subset M_2 \subset \cdots \subset M_n ⊂ \cdots$$ be a chain of submodules of $M$. It follows that the subset $M_0 := \bigcup^\infty_{i=1} M_i \subset M$ is a submodule. By $(1)$, it is finitely generated, so we can write $M' = \sum^n_{i=1} R x_i$ for some $x_i \in M'$. By definition, $x_i \in M_{j_{i}}$ some $j_i$. Let $s$ be the maximum of the finitely many $j_i$s. Then $M' = M_s$. It follows that $M_s = M' = M_{s+i}$ for all $i \geq 0$.
$(2) \Rightarrow (3)$: Let $S$ be a non-empty set of submodules of $M$ and let $M_1$ lie in $S$. If $M_1$ is not maximal, there exists an $M_2 \in S$ with $M_1 < M_2$. Inductively, if $M_i$ is not maximal, there exists an $M_{i+1}$ in $S$ with $Mi < M_{i+1}$. By the ascending chain condition, the sequence $$M_1 < M_2 < \dots < M_i < \cdots$$ must terminate. [Yeah I know, I used the axiom of choice, I'm not afraid of it.]
$(3) \Rightarrow (1)$: Let $N \subset M$ be a submodule and set $$S := \{ M_i \mid M_i \subset N \text{ is a finitely generated submodule} \}.$$ Then $(0) \in S$ so $S \neq \phi$. By assumption, there exists a maximal element $M' \in S$. If $N \neq M'$, then there exists $x \in N \ M'$. But $M'$ finitely generated means that $M' + Rx \subset N$ is also finitely generated, so the submodule $M' + Rx$ of $N$ lies in $S$. This contradicts the maximality of $M'$. Hence $N = M'$ is finitely generated. $\square$

Remark 2. Let $R = F[t_1, \dots , t_n, \dots]$ (infinitely many $t_i$). Let $M = R$ as an $R$-module. Then $M$ is finitely generated since it is cyclic, but the ideal $(t_1, \dots , t_n, \dots)$ is clearly not finitely generated, and so $R$ is not a Noetherian $R$-module. Thus, in general, submodules of finitely generated modules need not be finitely generated.

Definition 3. Let $R$ be a commutative ring. We say that $R$ is a Noetherian ring if $R$ is a Noetherian $R$-module.

Proposition 4. Let $M$ be an $R$-module and $N$ a submodule of $M$. Then $M$ is $R$-Noetherian if and only if $N$ and $M/N$ are $R$-Noetherian. In particular, if $$0 \to M' \to M \to M'' → 0$$ is an exact sequence of $R$-modules with two of the modules $M$, $M'$, $M''$ being $R$-Noetherian, then they all are $R$-Noetherian.

Proof. $(\Rightarrow)$: Since $N' \subset N$ is a submodule, then $N' ⊂ M$ is a submodule hence finitely generated, or any ascending chain in $N$ is an ascending chain in $M$. Thus $N$ is $R$-Noetherian. By the Correspondence Principle, a (countable) chain of submodules in $M/N$ has the form $M_1/N \subset M_2/N \subset \cdots$ where $N \subset M_1 \subset M_2 \subset \cdots$ is a chain of submodules of $M$. Thus there exists an $r$ such that $M_r = M_{r+j}$ for all $j \geq 0$ and hence $M_r/N = M_{r+j}/N$ for all $j \geq 0.$
$(\Leftarrow)$: Let $M_1 \subset M_2 \subset \cdots \subset M_n ⊂ \cdots$ be an ascending chain of submodules of $M$. Then, consider the projection map $p: M \to M/N$. Then, $p(M_n)$ is an ascending chain of $M/N$, so there exists $n_0: n > m > N_0$ implies $p(M_n) = p(M_m)$. There exists also $n_0'$ such that $n > m > n_0'$ implies that $M_n\cap N = M_m\cap N$.
Let $n>m>n_0$, $n_0'$. Let $x \in M_n$, then there exists $y \in M_m$ such that $p(y) = p(x)$ since $p(M_n) = p(M_m)$, $p(x-y) = 0$, which implies that $x-y \in N$, but $x-y\in M_n$ since $y \in M_m\subset M_n$, thus $x-y\in M_n\cap N=M_m\cap N$, thus $x-y\in M_m$, this implies $x\in M_m$. $\square$

Corollary 5. If $M$, $N$ are Noetherian $R$-modules, so is $M \sqcup N$.
Proof. $(M \sqcup N)/N \cong M$ and $N$ are Noetherian. $\square$

Theorem 6. Let $R$ be a Noetherian ring. If $M$ is a finitely generated $R$-module, then $M$ is $R$-Noetherian.
Proof. Suppose $M = \sum^n_{i=1} Rx_i$. Let $f : R^n \to M$ be the $R$-epimorphism given by $e_i \mapsto x_i$, where $\{e_1, \dots , e_n\}$ is the standard basis for $R^n$. Since $R$ is $R$-Noetherian so is $R_n$ by Corollary 5, and hence so is $M \cong R_n/ \ker f$ by Proposition 4. $\square$

Corollary 7. Let $R$ be a Noetherian ring and $M$ a finitely generated $R$-module. Then there exists positive integers $m$ and $n$ and an exact sequence $$R^m \xrightarrow{g} R^n \xrightarrow{f} M \to 0.$$

Proof. As in Theorem 6, there exists an integer $n$ and an $R$-epimorphism $f : R_n \to M$. Since $R^n$ is finitely generated, and hence $R$-Noetherian, $\ker f$ is also finitely generated. Therefore, there exists an $R$-epimorphism $h : R^m → \ker f$. Setting $g$ to be the composition of $f$ and the inclusion yields the result. $\square$

Proposition 8. Suppose that $f : R \to S$ is a ring epimorphism of commutative rings. If $R$ is a Noetherian ring, then $S$ is also a Noetherian ring.
Proof. Let $\mathfrak{A} \subset S$ be an ideal. Then $f^{−1}(\mathfrak{A}) \subset R$ is an ideal, and hence finitely generated. Thus $\mathfrak{A} = f(f^{−1}(\mathfrak{A}))$ is finitely generated. $\square$

Theorem 9. (Hilbert's Basis Theorem) If $R$ is a Noetherian ring, then so is the ring $R[t_1, \dots , t_n]$.
Proof. By induction on $n$, it suffices to show that $R[t]$ is Noetherian. Let $\mathfrak{B} \subset R[t]$ be an ideal. We must show that $\mathfrak{B}$ is finitely generated. Let $$\mathfrak{A} = \{r \in R \mid r = \text{ lead } f, f \in \mathfrak{B} \}.$$($\text{ lead }f$ is the leading coefficient of $f$.)
We first show that $\mathfrak{A}$ is an ideal. (Note that $0 \in \mathfrak{A}$ as $0 \in \mathfrak{B}$.) Let $a, b \in \mathfrak{A}$ and $r \in R$. We have to show $ra + b$ lies in $\mathfrak{A}$. We may assume that $a$ and $b$ are nonzero. Choose $f$, $g \in B$ say of degrees $m$ and $n$ respectively satisfying $\text{ lead }f = a$ and $\text{lead } g = b$. Set $h = rt^nf + t^mg$ in $\mathfrak{B}$. Then $ra + b = \text{lead } h$, proving $\mathfrak{A}$ is an ideal. As $R$ is Noetherian, $\mathfrak{A} = (a_1, \dots, a_n)$ for some $a_1, \dots , a_n \in \mathfrak{A}$ with $n \in \mathbb{Z}^+$. Choose $f_{id_i}$ in $\mathfrak{B}$ such that $a_i = \text{lead } f_{id_i}$ and $\deg f_{id_i} = d_i$. Let $\mathfrak{B}_0 = (f_{1d_1}, \dots , f_{nd_n})$, an ideal in $R[t]$, and $N = \max\{d_1, \dots, d_n\}$.
Let $f \in \mathfrak{B}$ with $\text{lead }f = a$ and $\deg f = d$. Suppose that $d > N$. There exist $r_i \in R$ satisfying $a = \sum^n_{i=1} r_ia_i$, hence $f − \sum^n_{i=1} r_it^{d−d_i}f_{id_i}$ lies in $\mathfrak{B}$ and has degree less than $d$. It follows by induction that there exists a $g \in \mathfrak{B}_0$ such that $f − g$ lies in $\mathfrak{B}$ with $\deg(f − g) \leq N$. As the $R$-module $M := \sum^N_{i=0} Rt^i$ is finitely generated and $R$ is Noetherian, $M$ is a Noetherian $R$-module. In particular, the submodule $M_0 =
\{f \in \mathfrak{B} \mid \deg f \leq N \}$ is finitely generated. If $M_0 = \sum^m_{i=0} R[t]g_i$, then $\mathfrak{B} = \mathfrak{B}_0 + \sum^m_{i=0} R[t]g_i$ is finitely generated. $\square$

Definition 10. Let $R \subset S$ be commutative rings. We say that $S$ is a finitely generated commutative $R$-algebra (or an affine $R$-algebra when $R$ is a field) if there exist $x_1, \dots, x_n$ in $S$ satisfying $S = R[x_1, \dots , x_n]$ as rings.

Proposition 11 Let $R$ be a commutative ring and $S$ a finitely generated commutative $R$-algebra. If $R$ is Noetherian, then so is $S$.

Proof. By Hilbert's Basis Theorem $R[x_1, \dots, x_n]$ is finitely generated. The result then follows by the Correspondence Principle. $\square$

Lemma 12. (Artin-Tate) Let $T$ be a commutative ring and $R \subset S$ be subrings of $T$. Suppose that $R$ is Noetherian and $T$ is a finitely generated commutative $R$-algebra. Suppose that as an $S$-module $T$ is finitely generated. Then $S$ is a finitely generated commutative $R$-algebra.

Proof. Let $T = R[x_1, \dots, x_n]$ as a commutative $R$-algebra for some $x_i \in T$, some $n$ and $T = \sum^m_{i=1}Sy_i$ as an $S$-module for some $y_i \in T$, some $m$. Then for all $i = 1, \dots , n$ and $p, q = 1, \dots , m$, we have equations: $$(i) \hspace5pt x_i = \sum^m_{j=1}a_{ij}y_j
\text{ for some } a_{ij} ∈ S$$ and $$(ii) \hspace5pt y_py_q = \sum^m_{k=1} b_{pqk}y_k \text{ for some } b_{pqk} \in S \text{ (since $T$ is a ring)}.$$ Let $S_0 = R[a_{ij}, b_{pqk} \mid i = 1, \dots , n \text{ and } j, p, q, k = 1, \dots , m]$, a finitely generated $R$-algebra. Then $R \subset S_0 \subset S \subset T$.
Claim. $T$ is a finitely generated $S_0$-module.
Let $f \in T$, so $f = \sum c_{i_1,\dots, i_n} x^{i_1}_1 \dots x^{i_n}_n$ for some $c_{i_1,\dots, i_n} \in R$. Applying properties $(i)$ and $(ii)$ repeatedly shows that $f \in \sum^m_{i=1}S_0y_i$. Consequently, $T = S_0y_1 + \cdots + S_0y_m$ as claimed. Since $S_0$ is a finitely generated commutative $R$-algebra, it is Noetherian by Proposition 11. Thus $T$, being a finitely generated $S_0$-module, is a Noetherian $S_0$-module. Consequently, $S \subset T$ is a finitely generated $S_0$-module. It follows immediately that $S$ is a finitely generated commutative $R$-algebra. $\square$

Lemma 13. Suppose that $L \subset K \subset M$ are fields, then $M$ is a finite dimensional vector space over $L$ if and only if both $M$ is a finite dimensional vector space over $K$ and $K$ is a finite dimensional vector space over $L$.

Lemma 14. Let $L \subset M$ be fields. Suppose that $x \in M$ has the property that $L[x]$ is not a finite dimensional vector space over $L$. Then $L[x] \cong L[t]$ as rings and $L(x) \cong L(t)$ as fields.

Lemma 15. (Zariski’s Lemma) Let $F$ be a field and $E$ a field containing $F$ such that $E$ is an affine $F$-algebra. Then $E$ is a finite dimensional vector space over $F$ (i.e., a finitely generated $F$-module).

Proof. Suppose that $E = F[x_1, \dots, x_m]$. Since $E$ is a field, $E = F(x_1, \dots , x_m)$. Suppose that $E$ is not a finite dimensional vector space over $F$. Using Lemma 13, we see that $F(x_i)$ is an infinite dimensional vector space over $F$ for some $i$. Relabeling, we may assume $i = 1$. Continuing in this way, we see that after relabeling the $x_i$, we may assume that $F(x_1, \dots , x_i)$ is not a finite dimensional vector space over $F(x_1, \dots, x_{i−1})$ for $1 \leq i \leq r$, some $r$, and $E$ is a finite dimensional vector space over $F(x_1, \dots , x_r)$.
Let $K = F(x_1, \dots , x_r)$. We have $E = K(x_{r+1}, \dots , x_m)$ is a finitely generated $K$-module and $F$ is a Noetherian ring (since it is a field). Thus, the Artin-Tate Lemma implies that $K$ is a finitely generated $F$-algebra.
Write $K = F[y_1, \dots , y_n]$ for some $y_i \in K$. By Lemma 14, it follows that and $F[x_1, \dots , x_r] \cong F[t_1, \dots , t_r]$ and $K \cong F(t_1, \dots , t_r)$ as rings. Thus we can write
$$y_i = \frac{f_i(x_1, \dots, x_r)}{g_i(x_1, \dots, x_r)}$$ for some $f_i$, $g_i \in F[t_1, \dots , t_r]$, $g_i \neq 0$, $1 \leq i \leq n$. Let $g = g_1 \cdot \cdots \cdot g_n$ in $F[t_1, \dots , t_r]$. Then, $g(x_1, \dots , x_r) \in F[x_1, \dots , x_r]$. We know the UFD $F[t_1, \dots , t_r]$ contains infinitely many non-associative irreducibles. Hence, in particular, there exists an irreducible $f ∈ F[t_1, \dots , t_r]$ such that $f \nmid g$. Thus $f(x_1, \dots , x_r) \nmid g(x_1, \dots , x_r)$ in $F[x_1, \dots , x_r]$. As $K$ is a field, $$ \frac{1}{f(x_1, \dots , x_r)} \in K = F[y_1, \dots , y_n] = F\left[
\frac{f_1(x_1, \dots , x_r)}{g_1(x_1, \dots , x_r)}, \dots, \frac{f_n(x_1, \dots , x_r)}{g_n(x_1, \dots , x_r)}\right]. $$ This leads to the equation for $\frac{1}{f(x_1, \dots , x_r)}$. Choosing an appropriate $N \geq 0$ to clear denominators, we see that we have an equation $$ \frac{g(x_1, \dots , x_r)^N}{f(x_1, \dots , x_r)} ∈ F[f_1(x_1, \dots , x_r), \dots, f_n(x_1, \dots , x_r)]. $$ Then $\frac{g(x_1, \dots , x_r)^N}{f(x_1, \dots , x_r)}$ lies in $F[x_1, \dots , x_r]$, i.e., $f \mid g^N$ in $F[t_1, \dots , t_r]$, which is a contradiction. $\square$

Note. If $R = F[t_1, \dots , t_n]$ and $\mathfrak{A} \subset R$ is an ideal, the affine variety of $\mathfrak{A}$ in $F^n$ is defined by $$Z_F (\mathfrak{A}) = \{a = (a_1, \dots , a_n) \in F^n \mid f(\underline a) = 0 \text{ for all } f \in \mathfrak{A}\}.$$By Hilbert's Basis Theorem, there exist $f_1, \dots , f_r$ in $F[t_1, \dots , t_n]$ satisfying $\mathfrak{A} = (f_1, \dots , f_r)$. We then also write $Z_F(f_1, \dots , f_r)$ for $Z_F (\mathfrak{A})$. In particular, $\underline a$ lies in $Z_F (\mathfrak{A})$ if and only if $f_i(\underline a) = 0$ for $i = 1, \dots , r$, as any element in $\mathfrak{A}$ is an $R$-linear combination of the $f_i$’s.

Theorem 16. (Hilbert's Nullstellensatz) (Weak Form) Suppose $F$ is an algebraically closed field, $R = F[t_1, \dots , t_n]$, and $\mathfrak{A} = (t_1, \dots , t_r)$ is an ideal in $R$. Then $Z_F (\mathfrak{A})$ is the empty set if and only if $A$ is the unit ideal. In particular, if $\mathfrak{A} < R$, then there exists a point $\underline a \in F^n$ satisfying $f_1(\underline a) = 0, \dots, f_r(\underline a) = 0$.
Proof. Certainly if $\mathfrak{A} = R$, then there exists no $\underline a \in F^n$ such that the element $1$ in $R$ evaluated at $\underline a$ takes the value zero, so we need only show if $\mathfrak{A} < R$, then $Z_F (\mathfrak{A})$ is not empty.
So suppose that $\mathfrak{A} < R$. Then there exists a maximal ideal $\mathfrak{m} < R$ satisfying $\mathfrak{A} \subset \mathfrak{m}$. Let $\bar{} : R \to R/\mathfrak{m}$ be the canonical ring epimorphism. Set $E = R/\mathfrak{m} = F[\bar t_1, \dots , \bar t_n]$. As $R$ is an affine $F$-algebra, so is $E$. By Zariski’s Lemma, $E$ is a finite dimensional $F$-vector space.
Claim: $E = F$.
Indeed let $x \in E$. Then $F[x]$ is a finite dimensional $F$-vector space, so $\{1, x, x^2 , \dots, x^n\}$ must be linear dependent over $F$ for some $n$, i.e., $x$ is a root of some non-zero polynomial $f \in F[t]$. But any such $f$ factors completely over $F$, since $F$ is algebraically closed. Therefore, $E = F$. Consequently the point $\underline t = (\bar t_1, \dots , \bar t_n)$ in $E^n = F^n$ lies in $Z_F (\mathfrak{A})$, since $\mathfrak{A} \subset \mathfrak{m}$, i.e., $\bar{\mathfrak{A}} = \bar{\mathfrak{m}} = 0$ $\square$

What the Weak Form of Hilbert's Nullstellensatz says is that the maximal ideals in the ring of continuous real-valued functions on a finite and closed (or compact) set shall be in one-to-one correspondence with the points of the set over an algebraically closed field. The geometric interpretation of the Weak Hilbert Nullstellensatz is that the intersection of the “hyperplanes”
$$f_1 = 0, \dots , f_r = 0 \text { in } F^n$$ always contains a common point over an algebraically closed field $F$, unless the $f_i$ generate the unit ideal in $F[t_1, \dots , t_n]$.
I hope glimpses of the link between algebra and geometry are starting to glean through. In the next post I shall prove the stronger form of the Nullstellensatz, and introduce the spectrum $\text{Spec}(R)$ of a ring $R$.