Difference between revisions of "Gauss's Lemma (polynomial)"
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Suppose now that <math>I(P) = I(Q) = (1)</math>, and suppose <math>I(PQ) \neq (1)</math>. Then by [[Krull's Theorem]], there is a (proper) maximal ideal <math>\mathfrak{m}</math> that contains <math>I(PQ)</math>. Let | Suppose now that <math>I(P) = I(Q) = (1)</math>, and suppose <math>I(PQ) \neq (1)</math>. Then by [[Krull's Theorem]], there is a (proper) maximal ideal <math>\mathfrak{m}</math> that contains <math>I(PQ)</math>. Let | ||
| − | <cmath> P = \sum_{k \ge 0} p_k x^k | + | <cmath> P = \sum_{k \ge 0} p_k x^k, \qquad Q = \sum_{k\ge 0} q_k x^k . </cmath> |
Since <math>\mathfrak{m}</math> is maximal, there exist least positive integers <math>a,b</math> such that <math>p_a , q_b \notin \mathfrak{m}</math>. Then the coefficient of <math>x^{a+b}</math> of <math>PQ</math>, | Since <math>\mathfrak{m}</math> is maximal, there exist least positive integers <math>a,b</math> such that <math>p_a , q_b \notin \mathfrak{m}</math>. Then the coefficient of <math>x^{a+b}</math> of <math>PQ</math>, | ||
<cmath> \sum_{k=0}^{a+b} p_k q_{a+b-k} \equiv p_ap_b \pmod{\mathfrak{m}} , </cmath> | <cmath> \sum_{k=0}^{a+b} p_k q_{a+b-k} \equiv p_ap_b \pmod{\mathfrak{m}} , </cmath> | ||
Revision as of 18:12, 21 June 2017
Gauss's Lemma for Polynomials is a result in algebra.
The original statement concerns polynomials with integer coefficients. Such a polynomial is called primitive if the greatest common divisor of its coefficients is 1. The original lemma states that the product of two polynomials with integer coefficients is primitive if and only if each of the factor polynomials is primitive.
This result can be extended to more general settings. Specifically, let 
be a (not necessarily commutative) ring; let ![$A[[X]]$](http://latex.artofproblemsolving.com/2/d/6/2d6a0b62feae11d0bc2d413235381ed8e9d409a7.png)
be the ring of formal power series over . For ![$P \in A[[X]]$](http://latex.artofproblemsolving.com/5/9/d/59dc80cea132249c319a699f496971d438e34400.png)
, let 
denote the two-sided ideal generated by the coefficients of 
. The more general result states that for any ![$P,Q \in A[[X]]$](http://latex.artofproblemsolving.com/a/e/c/aecce3cdee303926289d2166222def03cef7e58c.png)
, 
if and only if 
and 
.
The original statement is a special case of the general statement, with 
, and and 
power series with finitely many nonzero coefficients.
Proof
It follows from definitions that 
; thus if , then 
. It remains to prove the converse.
Suppose now that , and suppose 
. Then by Krull's Theorem, there is a (proper) maximal ideal 
that contains 
. Let
![\[P = \sum_{k \ge 0} p_k x^k, \qquad Q = \sum_{k\ge 0} q_k x^k .\]](http://latex.artofproblemsolving.com/3/a/8/3a84872be9b0ec78c282259c7fcf772c602ecec5.png)
Since is maximal, there exist least positive integers 
such that 
. Then the coefficient of 
of 
,
![\[\sum_{k=0}^{a+b} p_k q_{a+b-k} \equiv p_ap_b \pmod{\mathfrak{m}} ,\]](http://latex.artofproblemsolving.com/2/e/6/2e6426ebe207baedfc3bdcec13cdf059fb72f625.png)
is not an element of , since is a maximal, and therefore a prime, ideal. This is a contradiction. Therefore 
if . 
Generalizations
One might be tempted to come to the more general conclusion that 
, but this is false. For instance, if 
are indeterminates, then the ideal generated by the coefficients of
![\[(ac)x^2 + (ad + bc)x + bd = (ax+b)(cx+d)\]](http://latex.artofproblemsolving.com/0/3/b/03b06463698b2a9c6636f2b87e13fa16f6661f49.png)
is a proper subset of 
.
Formal power series over a family of variables? I think the result is still true, and the proof would follow along similar lines, but it would be more involved.
