The Stacks project

Proof. The statement on localizations follows from the fact that any ideal $J \subset S^{-1}R$ is of the form $I \cdot S^{-1}R$. Any quotient $R/I$ of a Noetherian ring $R$ is Noetherian because any ideal $\overline{J} \subset R/I$ is of the form $J/I$ for some ideal $I \subset J \subset R$. Thus it suffices to show that if $R$ is Noetherian so is $R[X]$. Suppose $J_1 \subset J_2 \subset \ldots $ is an ascending chain of ideals in $R[X]$. Consider the ideals $I_{i, d}$ defined as the ideal of elements of $R$ which occur as leading coefficients of degree $d$ polynomials in $J_ i$. Clearly $I_{i, d} \subset I_{i', d'}$ whenever $i \leq i'$ and $d \leq d'$. By the ascending chain condition in $R$ there are at most finitely many distinct ideals among all of the $I_{i, d}$. (Hint: Any infinite set of elements of $\mathbf{N} \times \mathbf{N}$ contains an increasing infinite sequence.) Take $i_0$ so large that $I_{i, d} = I_{i_0, d}$ for all $i \geq i_0$ and all $d$. Suppose $f \in J_ i$ for some $i \geq i_0$. By induction on the degree $d = \deg (f)$ we show that $f \in J_{i_0}$. Namely, there exists a $g\in J_{i_0}$ whose degree is $d$ and which has the same leading coefficient as $f$. By induction $f - g \in J_{i_0}$ and we win. $\square$

Proof. Since $R[[x_1, \ldots , x_{n + 1}]] \cong R[[x_1, \ldots , x_ n]][[x_{n + 1}]]$ it suffices to prove the statement that $R[[x]]$ is Noetherian if $R$ is Noetherian. Let $I \subset R[[x]]$ be a ideal. We have to show that $I$ is a finitely generated ideal. For each integer $d$ denote $I_ d = \{ a \in R \mid ax^ d + \text{h.o.t.} \in I\} $. Then we see that $I_0 \subset I_1 \subset \ldots $ stabilizes as $R$ is Noetherian. Choose $d_0$ such that $I_{d_0} = I_{d_0 + 1} = \ldots $. For each $d \leq d_0$ choose elements $f_{d, j} \in I \cap (x^ d)$, $j = 1, \ldots , n_ d$ such that if we write $f_{d, j} = a_{d, j}x^ d + \text{h.o.t}$ then $I_ d = (a_{d, j})$. Denote $I' = (\{ f_{d, j}\} _{d = 0, \ldots , d_0, j = 1, \ldots , n_ d})$. Then it is clear that $I' \subset I$. Pick $f \in I$. First we may choose $c_{d, i} \in R$ such that

Next, we can choose $c_{i, 1} \in R$, $i = 1, \ldots , n_{d_0}$ such that

Next, we can choose $c_{i, 2} \in R$, $i = 1, \ldots , n_{d_0}$ such that

And so on. In the end we see that

is contained in $I'$ as desired. $\square$

Proof. This is immediate from Lemma 10.31.1 and the fact that fields are Noetherian rings and that $\mathbf{Z}$ is Noetherian ring (because it is a principal ideal domain). $\square$

Proof. Let $M$ be a finite $R$-module. By Lemma 10.5.4 we can find a finite filtration of $M$ whose successive quotients are of the form $R/I$. Since any ideal is finitely generated, each of the quotients $R/I$ is finitely presented. Hence $M$ is finitely presented by Lemma 10.5.3. This proves (1).

Let $N \subset M$ be a submodule. As $M$ is finite, the quotient $M/N$ is finite. Thus $M/N$ is of finite presentation by part (1). Thus we see that $N$ is finite by Lemma 10.5.3 part (5). This proves part (2).

To see (3) note that any ideal of $R[x_1, \ldots , x_ n]$ is finitely generated by Lemma 10.31.1. $\square$

Proof. This is because any closed subset of $\mathop{\mathrm{Spec}}(R)$ is uniquely of the form $V(I)$ with $I$ a radical ideal, see Lemma 10.17.2. And this correspondence is inclusion reversing. Thus the result follows from the definitions. $\square$

Proof. By Lemma 10.31.5 and Topology, Lemma 5.9.2 we see there are finitely many irreducible components. By Lemma 10.26.1 these correspond to minimal primes of $R$. $\square$

Proof. By Lemma 10.14.2 finite type is stable under base change. Thus $S \to S'$ is of finite type. Since $S$ is Noetherian we can apply Lemma 10.31.1. $\square$

Proof. Since $K/k$ is a finitely generated field extension, there exists a finitely generated $k$-algebra $B \subset K$ such that $K$ is the fraction field of $B$. In other words, $K = S^{-1}B$ with $S = B \setminus \{ 0\} $. Then $K \otimes _ k R = S^{-1}(B \otimes _ k R)$. Then $B \otimes _ k R$ is Noetherian by Lemma 10.31.7. Finally, $K \otimes _ k R = S^{-1}(B \otimes _ k R)$ is Noetherian by Lemma 10.31.1. $\square$

Proof. If $R$ is a domain, then $R \subset R_\mathfrak p$, hence $f = 1$ works. If $R$ is Noetherian, then the kernel $I$ of $R \to R_\mathfrak p$ is a finitely generated ideal and we can find $f \in R$, $f \not\in \mathfrak p$ such that $IR_ f = 0$. For this $f$ the map $R_ f \to R_\mathfrak p$ is injective and $f$ works. If $R$ is reduced with finitely many minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ n$, then we can choose $f \in \bigcap _{\mathfrak p_ i \not\subset \mathfrak p} \mathfrak p_ i$, $f \not\in \mathfrak p$. Indeed, if $\mathfrak {p}_ i\not\subset \mathfrak {p}$ then there exist $f_ i \in \mathfrak {p}_ i$, $f_ i \not\in \mathfrak {p}$ and $f = \prod f_ i$ works. For this $f$ we have $R_ f \subset R_\mathfrak p$ because the minimal primes of $R_ f$ correspond to minimal primes of $R_\mathfrak p$ and we can apply Lemma 10.25.2 (some details omitted). $\square$

Proof. If $f : R \to R$ were such an endomorphism but not injective, then

would be a strictly increasing chain of ideals. $\square$