The Stacks project

Proof. If $x = uy$ for some unit $u \in R$, then $(x) \subset (y)$ and $y = u^{-1}x$ so also $(y) \subset (x)$. Conversely, suppose that $(x) = (y)$. Then $x = fy$ and $y = gx$ for some $f, g \in A$. Then $x = fg x$ and since $R$ is a domain $fg = 1$. Thus $x$ and $y$ are associates. $\square$

Proof. Let $x$ be a nonzero element, not a unit, which does not have a factorization into irreducibles. Set $x_1 = x$. We can write $x = yz$ where neither $y$ nor $z$ is irreducible or a unit. Then either $y$ does not have a factorization into irreducibles, in which case we set $x_2 = y$, or $z$ does not have a factorization into irreducibles, in which case we set $x_2 = z$. Continuing in this fashion we find a sequence

of elements of $R$ with $x_ n/x_{n + 1}$ not a unit. This gives a strictly increasing sequence of principal ideals $(x_1) \subset (x_2) \subset (x_3) \subset \ldots$ thereby finishing the proof. $\square$

Proof. Assume $R$ is a UFD and let $x \in R$ be an irreducible element. Say $ab \in (x)$, i.e., $ab = cx$. Choose factorizations $a = a_1 \ldots a_ n$, $b = b_1 \ldots b_ m$, and $c = c_1 \ldots c_ r$. By uniqueness of the factorization

we find that $x$ is an associate of one of the elements $a_1, \ldots , b_ m$. In other words, either $a \in (x)$ or $b \in (x)$ and we conclude that $x$ is prime.

Assume every irreducible element is prime. We have to prove that factorization into irreducibles is unique up to permutation and taking associates. Say $a_1 \ldots a_ m = b_1 \ldots b_ n$ with $a_ i$ and $b_ j$ irreducible. Since $a_1$ is prime, we see that $b_ j \in (a_1)$ for some $j$. After renumbering we may assume $b_1 \in (a_1)$. Then $b_1 = a_1 u$ and since $b_1$ is irreducible we see that $u$ is a unit. Hence $a_1$ and $b_1$ are associates and $a_2 \ldots a_ n = ub_2\ldots b_ m$. By induction on $n + m$ we see that $n = m$ and $a_ i$ associate to $b_{\sigma (i)}$ for $i = 2, \ldots , n$ as desired. $\square$

Proof. Assume $R$ is a UFD and let $\mathfrak p$ be a height 1 prime ideal. Take $x \in \mathfrak p$ nonzero and let $x = a_1 \ldots a_ n$ be a factorization into irreducibles. Since $\mathfrak p$ is prime we see that $a_ i \in \mathfrak p$ for some $i$. By Lemma 10.120.5 the ideal $(a_ i)$ is prime. Since $\mathfrak p$ has height $1$ we conclude that $(a_ i) = \mathfrak p$.

Assume every height $1$ prime is principal. Since $R$ is Noetherian every nonzero nonunit element $x$ has a factorization into irreducibles, see Lemma 10.120.3. It suffices to prove that an irreducible element $x$ is prime, see Lemma 10.120.5. Let $(x) \subset \mathfrak p$ be a prime minimal over $(x)$. Then $\mathfrak p$ has height $1$ by Lemma 10.60.11. By assumption $\mathfrak p = (y)$. Hence $x = yz$ and $z$ is a unit as $x$ is irreducible. Thus $(x) = (y)$ and we see that $x$ is prime. $\square$

Proof. Say $x = \alpha \beta$ for $\alpha , \beta \in S^{-1}A$. Then $\alpha = a/s$ and $\beta = b/s'$ for $a, b \in A$, $s, s' \in S$. Thus we get $ss'x = ab$. By assumption we can write $ss' = p_1 \ldots p_ r$ for some prime elements $p_ i$. For each $i$ the element $p_ i$ divides either $a$ or $b$. Dividing we find a factorization $x = a' b'$ and $a = s'' a'$, $b = s''' b'$ for some $s'', s''' \in S$. As $x$ is irreducible, either $a'$ or $b'$ is a unit. Tracing back we find that either $\alpha$ or $\beta$ is a unit. This proves (1).

Suppose $x$ is prime. Then $A/(x)$ is a domain. Hence $S^{-1}A/xS^{-1}A = S^{-1}(A/(x))$ is a domain or zero. Thus $x$ maps to a prime element or a unit.

Suppose that the image of $x$ in $S^{-1}A$ is a unit. Then $y x = s$ for some $s \in S$ and $y \in A$. By assumption $s = p_1 \ldots p_ r$ with $p_ i$ a prime element. For each $i$ either $p_ i$ divides $y$ or $p_ i$ divides $x$. In the second case $p_ i$ and $x$ are associates (as $x$ is irreducible) and we are done. But if the first case happens for all $i = 1, \ldots , r$, then $x$ is a unit which is a contradiction.

Suppose that the image of $x$ in $S^{-1}A$ is a prime element. Assume $a, b \in A$ and $ab \in (x)$. Then $sa = xy$ or $sb = xy$ for some $s \in S$ and $y \in A$. Say the first case happens. By assumption $s = p_1 \ldots p_ r$ with $p_ i$ a prime element. For each $i$ either $p_ i$ divides $y$ or $p_ i$ divides $x$. In the second case $p_ i$ and $x$ are associates (as $x$ is irreducible) and we are done. If the first case happens for all $i = 1, \ldots , r$, then $a \in (x)$ as desired. This completes the proof of (2).

The final statement of the lemma follows from (1) and (2) and Lemma 10.120.5. $\square$

Proof. Consider an ascending chain $(a_1) \subset (a_2) \subset (a_3) \subset \ldots$ of principal ideals in $R$. Write $a_1 = p_1^{e_1} \ldots p_ r^{e_ r}$ with $p_ i$ prime. Then we see that $a_ n$ is an associate of $p_1^{c_1} \ldots p_ r^{c_ r}$ for some $0 \leq c_ i \leq e_ i$. Since there are only finitely many possibilities we conclude. $\square$

Proof. Consider an ascending chain $(f_1) \subset (f_2) \subset (f_3) \subset \ldots$ of principal ideals in $R[x]$. Since $f_{n + 1}$ divides $f_ n$ we see that the degrees decrease in the sequence. Thus $f_ n$ has fixed degree $d \geq 0$ for all $n \gg 0$. Let $a_ n$ be the leading coefficient of $f_ n$. The condition $f_ n \in (f_{n + 1})$ implies that $a_{n + 1}$ divides $a_ n$ for all $n$. By our assumption on $R$ we see that $a_{n + 1}$ and $a_ n$ are associates for all $n$ large enough (Lemma 10.120.2). Thus for large $n$ we see that $f_ n = u f_{n + 1}$ where $u \in R$ (for reasons of degree) is a unit (as $a_ n$ and $a_{n + 1}$ are associates). $\square$

Proof. Let $R$ be a UFD. Then $R$ satisfies the ascending chain condition for principal ideals (Lemma 10.120.8), hence $R[x]$ satisfies the ascending chain condition for principal ideals (Lemma 10.120.9), and hence every element of $R[x]$ has a factorization into irreducibles (Lemma 10.120.3). Let $S \subset R$ be the multiplicative subset generated by prime elements. Since every nonunit of $R$ is a product of prime elements we see that $K = S^{-1}R$ is the fraction field of $R$. Observe that every prime element of $R$ maps to a prime element of $R[x]$ and that $S^{-1}(R[x]) = S^{-1}R[x] = K[x]$ is a UFD (and even a PID). Thus we may apply Lemma 10.120.7 to conclude. $\square$

Proof. Let $R$ be a UFD. Let $x$ be an element of the fraction field of $R$ which is integral over $R$. Say $x^ d - a_1 x^{d - 1} - \ldots - a_ d = 0$ with $a_ i \in R$. We can write $x = u p_1^{e_1} \ldots p_ r^{e_ r}$ with $u$ a unit, $e_ i \in \mathbf{Z}$, and $p_1, \ldots , p_ r$ irreducible elements which are not associates. To prove the lemma we have to show $e_ i \geq 0$. If not, say $e_1 < 0$, then for $N \gg 0$ we get

which contradicts uniqueness of factorization in $R$. $\square$

Proof. As a PID is Noetherian this follows from Lemma 10.120.6. $\square$

Proof. Let $R$ be a PID. Since every nonzero ideal of $R$ is principal, and $R$ is a UFD (Lemma 10.120.13), this follows from the fact that every irreducible element in $R$ is prime (Lemma 10.120.5) so that factorizations of elements turn into factorizations into primes. $\square$

Proof. It suffices to show that $I$ and $J$ are finite locally free $A$-modules of rank $1$, see Lemma 10.78.2. To do this, write $f = \sum _{i = 1, \ldots , n} x_ i y_ i$ with $x_ i \in I$ and $y_ i \in J$. We can also write $x_ i y_ i = a_ i f$ for some $a_ i \in A$. Since $f$ is a nonzerodivisor we see that $\sum a_ i = 1$. Thus it suffices to show that each $I_{a_ i}$ and $J_{a_ i}$ is free of rank $1$ over $A_{a_ i}$. After replacing $A$ by $A_{a_ i}$ we conclude that $f = xy$ for some $x \in I$ and $y \in J$. Note that both $x$ and $y$ are nonzerodivisors. We claim that $I = (x)$ and $J = (y)$ which finishes the proof. Namely, if $x' \in I$, then $x'y = af = axy$ for some $a \in A$. Hence $x' = ax$ and we win. $\square$

Proof. Assume (1). The argument is nontrivial because we did not assume that $R$ was Noetherian in our definition of a Dedekind domain. Let $\mathfrak p \subset R$ be a prime ideal. Observe that $\mathfrak p \not= \mathfrak p^2$ by uniqueness of the factorizations in the definition. Pick $x \in \mathfrak p$ with $x \not\in \mathfrak p^2$. Let $y \in \mathfrak p$ be a second element (for example $y = 0$). Write $(x, y) = \mathfrak p_1 \ldots \mathfrak p_ r$. Since $(x, y) \subset \mathfrak p$ at least one of the primes $\mathfrak p_ i$ is contained in $\mathfrak p$. But as $x \not\in \mathfrak p^2$ there is at most one. Thus exactly one of $\mathfrak p_1, \ldots , \mathfrak p_ r$ is contained in $\mathfrak p$, say $\mathfrak p_1 \subset \mathfrak p$. We conclude that $(x, y)R_\mathfrak p = \mathfrak p_1R_\mathfrak p$ is prime for every choice of $y$. We claim that $(x)R_\mathfrak p = \mathfrak pR_\mathfrak p$. Namely, pick $y \in \mathfrak p$. By the above applied with $y^2$ we see that $(x, y^2)R_\mathfrak p$ is prime. Hence $y \in (x, y^2)R_\mathfrak p$, i.e., $y = ax + by^2$ in $R_\mathfrak p$. Thus $(1 - by)y = ax \in (x)R_\mathfrak p$, i.e., $y \in (x)R_\mathfrak p$ as desired.

Writing $(x) = \mathfrak p_1 \ldots \mathfrak p_ r$ anew with $\mathfrak p_1 \subset \mathfrak p$ we conclude that $\mathfrak p_1 R_\mathfrak p = \mathfrak p R_\mathfrak p$, i.e., $\mathfrak p_1 = \mathfrak p$. Moreover, $\mathfrak p_1 = \mathfrak p$ is a finitely generated ideal of $R$ by Lemma 10.120.16. We conclude that $R$ is Noetherian by Lemma 10.28.10. Moreover, it follows that $R_\mathfrak m$ is a discrete valuation ring for every prime ideal $\mathfrak p$, see Lemma 10.119.7.

The equivalence of (2) and (3) follows from Lemmas 10.37.10 and 10.119.7. Assume (2) and (3) are satisfied. Let $I \subset R$ be an ideal. We will construct a factorization of $I$. If $I$ is prime, then there is nothing to prove. If not, pick $I \subset \mathfrak p$ with $\mathfrak p \subset R$ maximal. Let $J = \{ x \in R \mid x \mathfrak p \subset I\}$. We claim $J \mathfrak p = I$. It suffices to check this after localization at the maximal ideals $\mathfrak m$ of $R$ (the formation of $J$ commutes with localization and we use Lemma 10.23.1). Then either $\mathfrak p R_\mathfrak m = R_\mathfrak m$ and the result is clear, or $\mathfrak p R_\mathfrak m = \mathfrak m R_\mathfrak m$. In the last case $\mathfrak p R_\mathfrak m = (\pi )$ and the case where $\mathfrak p$ is principal is immediate. By Noetherian induction the ideal $J$ has a factorization and we obtain the desired factorization of $I$. We omit the proof of uniqueness of the factorization. $\square$

Proof. By Krull-Akizuki (Lemma 10.119.12) the ring $B$ is Noetherian. By Lemma 10.112.4 $\dim (B) = 1$. Thus $B$ is a Dedekind domain by Lemma 10.120.17. Surjectivity of the map on spectra follows from Lemma 10.36.17. The last two statements follow from Lemma 10.119.10. $\square$