# Content (polynomial)

The content of a polynomial over a ring is the greatest common divisor (in ) of the coefficients of the polynomial. The dependence on the ring is essential. ${\ displaystyle R}$${\ displaystyle R}$

This term has an application in Gauss's theorem. This represents the content of a product of two polynomials in relation to the content of its factors. This result is theoretically very interesting, since it can be used to prove that polynomial rings are factorial in finitely many variables over factorial rings , especially over bodies . In practice, the theorem can also be used to obtain restrictions on the rational zeros of a polynomial with whole coefficients. In particular, the candidates for rational zeros can be reduced to a finite number; this can be useful when factoring polynomials.

## definition

### For a factorial ring

Let be a polynomial with coefficients from any factorial ring . Then the content of and is referred to below with , whereby the English term is sometimes used in the literature . The content is clearly defined except for one unit . Next is determined. ${\ displaystyle \ textstyle f = \ sum _ {k = 0} ^ {n} a_ {k} x ^ {k} \ neq 0}$ ${\ displaystyle R}$${\ displaystyle \ operatorname {ggT} _ {R} (a_ {0}, \ dotsc, a_ {n})}$${\ displaystyle f}$${\ displaystyle \ operatorname {content} _ {R} (f)}$${\ displaystyle \ operatorname {cont} _ {R} (f)}$${\ displaystyle \ operatorname {content} _ {R} (0): = 0}$

### For the quotient field over a factorial ring

Let it be a factorial ring and the quotient field . The elements of the quotient field can be written using the prime elements as follows. ${\ displaystyle R}$${\ displaystyle K}$

${\ displaystyle a = ep_ {1} ^ {k_ {1}} p_ {2} ^ {k_ {2}} \ dotsm p_ {l} ^ {k_ {l}} \ in K ^ {*}}$with and pairwise not associated prime elements.${\ displaystyle k_ {1}, \ dotsc, k_ {l} \ in \ mathbb {Z}, \ e \ in R ^ {*}}$${\ displaystyle p_ {1}, \ dotsc, p_ {l} \ in R}$

The exponents that occur are uniquely determined and the evaluation can be made for each prime element${\ displaystyle p_ {1}}$

${\ displaystyle \ nu _ {p_ {1}} (a): = \ mathrm {ord} _ {p_ {1}} (a): = k_ {1}}$define with as above.${\ displaystyle a = ep_ {1} ^ {k_ {1}} p_ {2} ^ {k_ {2}} \ dotsm p_ {l} ^ {k_ {l}}}$

This allows the order for a polynomial to be determined with coefficients from the body . ${\ displaystyle K}$

${\ displaystyle \ mathrm {ord} _ {p} (f): = \ nu _ {p} (f): = \ min \ limits _ {i = 0, \ dotsc, n} \ nu _ {p} ( a_ {i})}$, where .${\ displaystyle f = \ sum _ {k = 0} ^ {n} a_ {k} x ^ {k} \ in K [X] \ backslash \ {0 \}}$

The content of can now also be defined using ${\ displaystyle f}$

${\ displaystyle \ mathrm {cont} _ {R} (f): = \ mathrm {inhalt} _ {R} (f): = \ prod _ {p \ in P} p ^ {\ mathrm {ord} _ { p} (f)}}$

A maximum amount of pairwise unassociated prime elements is assumed . For completeness one then defines ${\ displaystyle P}$${\ displaystyle R}$

${\ displaystyle \ nu _ {p} (0): = \ mathrm {ord} _ {p} (0): = \ infty}$ and ${\ displaystyle \ mathrm {cont} _ {R} (0): = \ mathrm {content} _ {R} (0): = 0}$

As in the case of a quotient field, the content is only unambiguously determined up to an association (another choice of leads to the multiplication of the content by a unit ). ${\ displaystyle P}$${\ displaystyle R}$

The two definitions are the same for polynomials over the ring , the second definition is a true generalization of the first. ${\ displaystyle R}$

If it is clear from which ring the coefficients of come from, one simply writes . ${\ displaystyle f}$${\ displaystyle \ mathrm {content} (f)}$

## Examples

Example 1 (for the 1st definition):

The contents of a polynomial with coefficients in is ${\ displaystyle f = 6x ^ {3} + 9x + 12}$${\ displaystyle \ mathbb {Z}}$

${\ displaystyle \ mathrm {inhalt} _ {\ mathbb {Z}} (f) = \ mathrm {ggT} _ {\ mathbb {Z}} (6,0,9,12) = 3}$

or also . On the other hand , if we understand it as a polynomial with coefficients , we get ${\ displaystyle -3}$${\ displaystyle f}$${\ displaystyle \ mathbb {Q}}$

${\ displaystyle \ mathrm {inhalt} _ {\ mathbb {Q}} (f) = \ mathrm {ggT} _ {\ mathbb {Q}} (6,0,9,12) = 1}$

or any other rational number besides zero.

Example 2 (for the 2nd definition):

The content of as a polynomial with coefficients from as a quotient field of is ${\ displaystyle f = x ^ {3} + {\ frac {3} {2}} x + 2}$${\ displaystyle K = \ mathbb {Q}}$${\ displaystyle R = \ mathbb {Z}}$

${\ displaystyle \ mathrm {contents} _ {\ mathbb {Z}} (f) = {\ frac {1} {2}}}$

or also . On the other hand , if we understand it as a polynomial with coefficients , we get ${\ displaystyle - {\ frac {1} {2}}}$${\ displaystyle f}$${\ displaystyle K = R = \ mathbb {Q}}$

${\ displaystyle \ mathrm {content} _ {\ mathbb {Q}} (f) = 1}$

or any other rational number besides zero.

## Remarks

Polynomials whose content is a unit are called primitive . With is the primitive proportion (engl. Primitive part ), respectively. ${\ displaystyle \ mathrm {pp} (f): = {\ tfrac {f} {\ mathrm {inhalt} _ {R} (f)}}}$

A polynomial with coefficients from the quotient field of a factorial ring is from the polynomial ring above if the content is in . ${\ displaystyle R}$${\ displaystyle R}$${\ displaystyle R}$

## Lemma of Gauss

### statement

Let it be a factorial ring and its quotient field , then applies to${\ displaystyle R}$${\ displaystyle K}$${\ displaystyle f, g \ in K [X]}$

${\ displaystyle \ mathrm {inhalt} _ {R} (f \ cdot g) = \ mathrm {inhalt} _ {R} (f) \ cdot \ mathrm {inhalt} _ {R} (g)}$,

in particular, the product of two primitive polynomials is primitive again.

### Corollaries

The following four corollaries from this statement are often referred to as Gauss's lemma :

• The polynomial ring over a factorial ring is factorial.${\ displaystyle R [X]}$${\ displaystyle R}$
• If a non-constant polynomial (in a variable) is irreducible over a factorial ring, then it is also irreducible over its quotient field.
• If a normalized polynomial has a zero in the quotient field, then this is already in the ring itself.
• The product of two normalized polynomials with rational coefficients only has integral coefficients if the coefficients of and are already integral.${\ displaystyle f, g}$${\ displaystyle f}$${\ displaystyle g}$

Further corollaries are:

• If a polynomial is given from the ring, each zero in the quotient field can be represented as a fraction in such a way that the denominator is a divisor of the highest coefficient and the numerator is a divisor of the absolute term (see also theorem on rational zeros ).
• The prime elements in the polynomial ring over a factorial ring are exactly the prime elements of the ring together with the primitive prime elements of the polynomial ring over the quotient field of .${\ displaystyle R [X]}$${\ displaystyle R}$${\ displaystyle K [X]}$${\ displaystyle K}$${\ displaystyle R}$
• If a factorial ring is, then the polynomial ring is factorial in finitely many variables .${\ displaystyle R}$${\ displaystyle R [X_ {1}, \ dotsc, X_ {n}]}$

### Proof idea

First you convince yourself that this is true for. So one can assume that are primitive (thus ), and thus only has to show this special case of the proposition. It's also easy to see that ${\ displaystyle f \ in K}$${\ displaystyle f, g}$${\ displaystyle f, g, fg \ in R [X]}$

${\ displaystyle \ mathrm {inhalt_ {R}} (f) = 1 \ Leftrightarrow \ forall p {\ text {Primelement}}: \ f + pR [X] \ neq 0 \ in R [X] / pR [X] \ cong (R / pR) [X]}$

But then the proposition is trivial, because and thus there is an integrity ring because is a prime ideal . ${\ displaystyle R / pR}$${\ displaystyle (R / pR) [X]}$${\ displaystyle pR}$

To the first corollary:

One proves that all prime elements of the ring and all primitive prime elements of prime are in. If you take advantage of the fact that the Euclidean ring is factorial, you can write each element out as the product of these prime elements (this had to be shown). The other corollaries do not need an idea of ​​proof. You just have to prove the statements directly. ${\ displaystyle K [X]}$${\ displaystyle R [X]}$${\ displaystyle K [X]}$${\ displaystyle R [X]}$

### Historical

Gauss himself shows the variant in the Disquisitiones Arithmeticae (art. 42):

• The product of two normalized polynomials with rational coefficients only has integral coefficients if the coefficients of and are already integral.${\ displaystyle f, g}$${\ displaystyle f}$${\ displaystyle g}$

### application

• ${\ displaystyle f = 6x ^ {7} + 23x ^ {3} + 9x ^ {2} -12x + 24}$is not divisible by in , because the content of is 1 and of 3.${\ displaystyle g = 6x ^ {4} + 9x ^ {3} + 3x ^ {2} + 3x + 3}$${\ displaystyle \ mathbb {Z} [X]}$${\ displaystyle f}$${\ displaystyle g}$
• ${\ displaystyle f = 2x ^ {7} + x ^ {3} + x ^ {2} + 3x + 1}$has no rational zeros, because the only possible rational zeros would be according to Gauss and .${\ displaystyle \ pm {\ tfrac {1} {2}}}$${\ displaystyle \ pm 1}$
• ${\ displaystyle f = 2x ^ {3} + 2x + 3}$is irreducible as a polynomial in , because it has degree 3 and no rational zeros (with Gauss you just have to check a finite number).${\ displaystyle \ mathbb {Q}}$
• ${\ displaystyle f = 8x ^ {5} -8x ^ {4} + 2x ^ {3} -8x ^ {2} -6x}$is to be factored as a polynomial in . The following trivial factorizations are first carried out (make them primitive and exclude with maximum power!):${\ displaystyle \ mathbb {C}}$${\ displaystyle x}$
${\ displaystyle f = 8x ^ {5} -8x ^ {4} + 2x ^ {3} -8x ^ {2} -6x = 2x (4x ^ {4} -4x ^ {3} + x ^ {2} -4x-3)}$
And so the remaining polynomial has the possible rational Gaussian zeros
${\ displaystyle \ pm 1, \ \ pm 3, \ pm {\ frac {1} {2}}, \ pm {\ frac {1} {4}}, \ pm {\ frac {3} {2}} , \ pm {\ frac {3} {4}}}$
By inserting it you can see that only and are the rational zeros. And by polynomial division we get ${\ displaystyle -0 {,} 5}$${\ displaystyle 1 {,} 5}$
${\ displaystyle f = 2x (x ^ {2} +1) \ left (x + {\ frac {1} {2}} \ right) \ left (x - {\ frac {3} {2}} \ right) = 2x (xi) (x + i) \ left (x + {\ frac {1} {2}} \ right) \ left (x - {\ frac {3} {2}} \ right)}$