Unique factorization ___domain: Difference between revisions

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== Counterexamples ==

:*The ring <math>\mathbb Z[\sqrt{-5}]</math> of all complex numbers of the form <math>a+ib\sqrt{5}</math>, where ''a'' and ''b'' are integers. Then 6 factors as both (2)(3) and as <math>\left(1+i\sqrt{5}\right)\left(1-i\sqrt{5}\right)</math>. These truly are different factorizations, because the only units in this ring are 1 and &minus;1; thus, none of 2, 3, <math>1+i\sqrt{5}</math>, and <math>1-i\sqrt{5}</math> are [[unit (ring theory)|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]].

Despite the examples given above, very few integral domains are UFDs. Here is a counterexample:

*Most [[ideal (ring theory)|factor ring]]s of a polynomial ring are not UFDs. Here is an example:

:The ring <math>\mathbb Z[\sqrt{-5}]</math> of all complex numbers of the form <math>a+ib\sqrt{5}</math>, where ''a'' and ''b'' are integers. Then 6 factors as both (2)(3) and as <math>\left(1+i\sqrt{5}\right)\left(1-i\sqrt{5}\right)</math>. These truly are different factorizations, because the only units in this ring are 1 and &minus;1; thus, none of 2, 3, <math>1+i\sqrt{5}</math>, and <math>1-i\sqrt{5}</math> are [[unit (ring theory)|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 [[ideal (ring theory)|factor ring]]s of a polynomial ring are not UFDs. Here is an example:

:Let <math>R</math> be any commutative ring. Then <math>R[X,Y,Z,W]/(XY-ZW)</math> is not a UFD. The proof is in two parts.

::First we must show <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> are all irreducible. Grade <math>R[X,Y,Z,W]/(XY-ZW)</math> by degree. Assume for a contradiction that <math>X</math> has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element <math>\alpha X + \beta Y + \gamma Z + \delta W</math> and a degree zero element <math>r</math>. This gives <math>X = r\alpha X + r\beta Y + r\gamma Z + r\delta W</math>. In <math>R[X,Y,Z,W]</math>, then, the degree one element <math>(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W</math> must be an element of the ideal <math>(XY-ZW)</math>, but the non-zero elements of that ideal are degree two and higher. Consequently, <math>(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W</math> must be zero in <math>R[X,Y,Z,W]</math>. That implies that <math>r\alpha = 1</math>, so <math>r</math> is a unit, which is a contradiction. <math>Y</math>, <math>Z</math>, and <math>W</math> are irreducible by the same argument.

::First we must show <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> are all irreducible. Grade <math>R[X,Y,Z,W]/(XY-ZW)</math> by degree. Assume for a contradiction that <math>X</math> has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element <math>\alpha X + \beta Y + \gamma Z + \delta W</math> and a degree zero element <math>r</math>. This gives <math>X = r\alpha X + r\beta Y + r\gamma Z + r\delta W</math>. In <math>R[X,Y,Z,W]</math>, then, the degree one element <math>(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W</math> must be an element of the ideal <math>(XY-ZW)</math>, but the non-zero elements of that ideal are degree two and higher. Consequently, <math>(r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W</math> must be zero in <math>R[X,Y,Z,W]</math>. That implies that <math>r\alpha = 1</math>, so <math>r</math> is a unit, which is a contradiction. <math>Y</math>, <math>Z</math>, and <math>W</math> are irreducible by the same argument.

::Next, the element <math>XY</math> equals the element <math>ZW</math> because of the relation <math>XY - ZW = 0</math>. That means that <math>XY</math> and <math>ZW</math> are two different factorizations of the same element into irreducibles, so <math>R[X,Y,Z,W]/(XY-ZW)</math> is not a UFD.

::Next, the element <math>XY</math> equals the element <math>ZW</math> because of the relation <math>XY - ZW = 0</math>. That means that <math>XY</math> and <math>ZW</math> are two different factorizations of the same element into irreducibles, so <math>R[X,Y,Z,W]/(XY-ZW)</math> is not a UFD.

== Properties ==