## 10.120 Factorization

Here are some notions and relations between them that are typically taught in a first year course on algebra at the undergraduate level.

Definition 10.120.1. Let $R$ be a domain.

1. Elements $x, y \in R$ are called associates if there exists a unit $u \in R^*$ such that $x = uy$.

2. An element $x \in R$ is called irreducible if it is nonzero, not a unit and whenever $x = yz$, $y, z \in R$, then $y$ is either a unit or an associate of $x$.

3. An element $x \in R$ is called prime if the ideal generated by $x$ is a prime ideal.

Lemma 10.120.2. Let $R$ be a domain. Let $x, y \in R$. Then $x$, $y$ are associates if and only if $(x) = (y)$.

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. $\square$

Lemma 10.120.3. Let $R$ be a domain. Consider the following conditions:

1. The ring $R$ satisfies the ascending chain condition for principal ideals.

2. Every nonzero, nonunit element $a \in R$ has a factorization $a = b_1 \ldots b_ k$ with each $b_ i$ an irreducible element of $R$.

Then (1) implies (2).

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

$x_1 | x_2 | x_3 | \ldots$

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. $\square$

Definition 10.120.4. A unique factorization domain, abbreviated UFD, is a domain $R$ such that if $x \in R$ is a nonzero, nonunit, then $x$ has a factorization into irreducibles, and if

$x = a_1 \ldots a_ m = b_1 \ldots b_ n$

are factorizations into irreducibles then $n = m$ and there exists a permutation $\sigma : \{ 1, \ldots , n\} \to \{ 1, \ldots , n\}$ such that $a_ i$ and $b_{\sigma (i)}$ are associates.

Lemma 10.120.5. Let $R$ be a domain. Assume every nonzero, nonunit factors into irreducibles. Then $R$ is a UFD if and only if every irreducible element is prime.

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

$a_1 \ldots a_ n b_1 \ldots b_ m = c_1 \ldots c_ r x$

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. $\square$

Lemma 10.120.6. Let $R$ be a Noetherian domain. Then $R$ is a UFD if and only if every height $1$ prime ideal is principal.

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. $\square$

Lemma 10.120.7 (Nagata's criterion for factoriality). Let $A$ be a domain. Let $S \subset A$ be a multiplicative subset generated by prime elements. Let $x \in A$ be irreducible. Then

1. the image of $x$ in $S^{-1}A$ is irreducible or a unit, and

2. $x$ is prime if and only if the image of $x$ in $S^{-1}A$ is a unit or a prime element in $S^{-1}A$.

Moreover, then $A$ is a UFD if and only if every element of $A$ has a factorization into irreducibles and $S^{-1}A$ is a UFD.

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. $\square$

Lemma 10.120.8. A UFD satisfies the ascending chain condition for principal ideals.

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. $\square$

Lemma 10.120.9. Let $R$ be a domain. Assume $R$ has the ascending chain condition for principal ideals. Then the same property holds for a polynomial ring over $R$.

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). $\square$

Lemma 10.120.10. A polynomial ring over a UFD is a UFD. In particular, if $k$ is a field, then $k[x_1, \ldots , x_ n]$ is a UFD.

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

$u^ d p_2^{de_2 + N} \ldots p_ r^{de_ r + N} = p_1^{-de_1}p_2^ N \ldots p_ r^ N( \sum \nolimits _{i = 1, \ldots , d} a_ i x^{d - i} ) \in (p_1)$

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

Definition 10.120.12. A principal ideal domain, abbreviated PID, is a domain $R$ such that every ideal is a principal ideal.

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

Definition 10.120.14. A Dedekind domain is a domain $R$ such that every nonzero ideal $I \subset R$ can be written as a product

$I = \mathfrak p_1 \ldots \mathfrak p_ r$

of nonzero prime ideals uniquely up to permutation of the $\mathfrak p_ i$.

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. $\square$

Lemma 10.120.16. Let $A$ be a ring. Let $I$ and $J$ be nonzero ideals of $A$ such that $IJ = (f)$ for some nonzerodivisor $f \in A$. Then $I$ and $J$ are finitely generated ideals and finitely locally free of rank $1$ as $A$-modules.

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. $\square$

Lemma 10.120.17. Let $R$ be a ring. The following are equivalent

1. $R$ is a Dedekind domain,

2. $R$ is a Noetherian domain, and for every maximal ideal $\mathfrak m$ the local ring $R_{\mathfrak m}$ is a discrete valuation ring, and

3. $R$ is a Noetherian, normal domain, and $\dim (R) \leq 1$.

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. $\square$

The following is a variant of the Krull-Akizuki lemma.

Lemma 10.120.18. Let $A$ be a Noetherian domain of dimension $1$ with fraction field $K$. Let $L/K$ be a finite extension. Let $B$ be the integral closure of $A$ in $L$. Then $B$ is a Dedekind domain and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective, has finite fibres, and induces finite residue field extensions.

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. $\square$

Comment #1695 by user26857 on

As someone noticed in http://math.stackexchange.com/questions/1537199/in-stacks-project-polynomial-ring-over-ufd-is-ufd, it seems the proof of Lemma 10.119.8 misses the part where one has to show that $R[X]$ is atomic, that is, every non-zero non-unit is a product of irreducibles.

Comment #1696 by user26857 on

Btw, why all results are called "Lemma"?

Comment #1697 by on

You can take a look at tag \tag{02BZ} where the philosophy regarding everything being a lemma (or sometimes proposition or theorem) is explained. So there are theorems and propositions, just not many of them (and what makes up the distinction might be vague at some points).

Comment #1703 by user26857 on

I don't think this "philosophy" is a good choice, but in the same time I don't care that much. Instead I'll be glad to see a complete proof of 10.119.8.

Comment #1743 by on

@#1695: OK, we've added the necessary lemmas. See here. Thanks!

Comment #3273 by Louis Martini on

There is a typo in the last paragraph of the proof of lemma 10.119.17. In one instance the ring $R$ is referred to as $A$.

Comment #6841 by David on

It appears that Lemma 0BC1 depends on a subsequence result, Lemma 034V

Comment #6977 by on

@#6841. Why? I think it relies on knowing that the polynomial ring in one variable over a field is both a UFD (and even a PID). Right?

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