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# définition - Unique_factorization_domain

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# Unique factorization domain

In mathematics, a unique factorization domain (UFD) is, roughly speaking, a commutative ring in which every element, with special exceptions, can be uniquely written as a product of prime elements (or irreducible elements), analogous to the fundamental theorem of arithmetic for the integers. UFDs are sometimes called factorial rings, following the terminology of Bourbaki.

Note that unique factorization domains appear in the following chain of class inclusions:

Commutative ringsintegral domainsintegrally closed domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfields

## Definition

Formally, a unique factorization domain is defined to be an integral domain R in which every non-zero and non-unit x of R can be written as a product (including an empty product) of irreducible elements pi of R and a unit u:

x = u p1 p2 ... pn with n≥0

and this representation is unique in the following sense: If q1,...,qm are irreducible elements of R and w is a unit such that

x = w q1 q2 ... qm with m≥0,

then m = n and there exists a bijective map φ : {1,...,n} -> {1,...,m} such that pi is associated to qφ(i) for i ∈ {1, ..., n}.

The uniqueness part is usually hard to verify, which is why the following equivalent definition is useful:

A unique factorization domain is an integral domain R in which every non-zero element can be written as a product of a unit and prime elements of R.

## Examples

Most rings familiar from elementary mathematics are UFDs:

Further examples of UFDs are:

• The formal power series ring K[[X1,...,Xn]] over a field K (or more generally over a PID but not over a UFD).

## Non-examples

• The quadratic integer ring $\mathbb Z[\sqrt{-5}]$ of all complex numbers of the form $a+ib\sqrt{5}$, where a and b are integers. Then 6 factors as both (2)(3) and as $\left(1+i\sqrt{5}\right)\left(1-i\sqrt{5}\right)$. These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, $1+i\sqrt{5}$, and $1-i\sqrt{5}$ are associate. It is not hard to show that all four factors are irreducible as well, though this may not be obvious. See also algebraic integer.
• Most factor rings of a polynomial ring are not UFDs. Here are two examples:
• Let $R$ be any commutative ring. Then $R[X,Y,Z,W]/(XY-ZW)$ is not a UFD. The proof is in two parts.
First we must show $X$, $Y$, $Z$, and $W$ are all irreducible. Grade $R[X,Y,Z,W]/(XY-ZW)$ by degree. Assume for a contradiction that $X$ has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element $\alpha X + \beta Y + \gamma Z + \delta W$ and a degree zero element $r$. This gives $X = r\alpha X + r\beta Y + r\gamma Z + r\delta W$. In $R[X,Y,Z,W]$, then, the degree one element $(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W$ must be an element of the ideal $(XY-ZW)$, but the non-zero elements of that ideal are degree two and higher. Consequently, $(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W$ must be zero in $R[X,Y,Z,W]$. That implies that $r\alpha = 1$, so $r$ is a unit, which is a contradiction. $Y$, $Z$, and $W$ are irreducible by the same argument.
Next, the element $XY$ equals the element $ZW$ because of the relation $XY - ZW = 0$. That means that $XY$ and $ZW$ are two different factorizations of the same element into irreducibles, so $R[X,Y,Z,W]/(XY-ZW)$ is not a UFD.
• The ring of holomorphic functions in a single complex variable is not a UFD, since there exist holomorphic functions with an infinity of zeros, and thus an infinity of irreducible factors, while a UFD factorization must be finite, e.g.:
$\sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1-{{z^2}\over{n^2}}\right)$.
• A Noetherian domain is not necessarily a UFD. Although any non-zero non-unit in a Noetherian domain is the product of irreducible elements, this product is not necessarily unique.

## Properties

Some concepts defined for integers can be generalized to UFDs:

• In UFDs, every irreducible element is prime. (In any integral domain, every prime element is irreducible, but the converse does not always hold.) Note that this has a partial converse: any Noetherian domain is a UFD if every irreducible element is prime.
• Any two (or finitely many) elements of a UFD have a greatest common divisor and a least common multiple. Here, a greatest common divisor of a and b is an element d which divides both a and b, and such that every other common divisor of a and b divides d. All greatest common divisors of a and b are associated.
• Let S be a multiplicaively closed subset of a UFD A. Then the localization $S^{-1}A$ is a UFD. A partial converse to this also holds; see below.

## Equivalent conditions for a ring to be a UFD

A Noetherian integral domain is a UFD if and only if every height 1 prime ideal is principal. Also, a Dedekind domain is a UFD if and only if its ideal class group is trivial. In this case it is in fact a principal ideal domain.

There are also equivalent conditions for non-noetherian integral domains. Let A be an integral domain. Then the following are equivalent.

1. A is a UFD.
2. Every nonzero prime ideal of A contains a prime element. (Kaplansky)
3. A satisfies ascending chain condition on principal ideals (ACCP), and the localization S−1A is a UFD, where S is a multiplicatively closed subset of A generated by prime elements. (Nagata criterion)
4. A satisfies (ACCP) and every irreducible is prime.
5. A is a GCD domain (i.e., any two elements have a greatest common divisor) satisfying (ACCP).
6. A is a Schreier domain,[1] and every nonzero nonunit can be expressed as a finite product of irreducible elements (that is, A is atomic.)
7. A has a divisor theory in which every divisor is principal.
8. A is a Krull domain in which every divisorial ideal is principal (in fact, this is the definition of UFD in Bourbaki.)
9. A is a Krull domain and every prime ideal of height 1 is principal.[2]

In practice, (2) and (3) are the most useful conditions to check. For example, it follows immediately from (2) that a PID is a UFD, since, in a PID, every prime ideal is generated by a prime element.

Let A be a Zariski ring (e.g., a local noetherian ring).If the completion $\widehat{A}$ is a UFD, then A is a UFD.[3]

## References

1. ^ A Schreier domain is an integrally closed integral domain where, whenever x divides yz, x can be written as x = x1 x2 so that x1 divides y and x2 divides z. In particular, a GCD domain is a Schreier domain
2. ^ Bourbaki, 7.3, no 2, Theorem 1.
3. ^ Bourbaki, 7.3, no 6, Proposition 4.

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