Difference between revisions of "Gauss's Lemma (polynomial)"

Revision as of 20:52, 21 September 2008

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 $A$ be a (not necessarily commutative) ring; let $A[[X]]$ be the ring of formal power series over $A$ . For $P \in A[[X]]$ , let $I(P)$ denote the two-sided ideal generated by the coefficients of $P$ . The more general result states that for any $P,Q \in A[[X]]$ , $I(PQ) = (1)$ if and only if $I(P) = (1)$ and $I(Q) = (1)$ .

The original statement is a special case of the general statement, with $A = \mathbb{Z}$ , and $P$ and $Q$ power series with finitely many nonzero coefficients.

Proof

It follows from definitions that $I(PQ) \subset I(P) I(Q)$ ; thus if $I(PQ) = (1)$ , then $I(P) = I(Q) = (1)$ . It remains to prove the converse.

Suppose now that $I(P) = I(Q) = (1)$ , and suppose $I(PQ) \neq (1)$ . Then by Krull's Theorem, there is a (proper) maximal ideal $\mathfrak{m}$ that contains $I(PQ)$ . Let

\[P = \sum_{k \ge 0} p_k x^k}, \qquad Q = \sum_{k\ge 0} q_k x^k .\] (Error compiling LaTeX. Unknown error_msg)

Since $\mathfrak{m}$ is maximal, there exist least positive integers $a,b$ such that $p_a , q_b \notin \mathfrak{m}$ . Then the coefficient of $x^{a+b}$ of $PQ$ , $\sum_{k=0}^{a+b} p_k q_{a+b-k} \equiv p_ap_b \pmod{\mathfrak{m}} ,$ is not an element of $\mathfrak{m}$ , since $\mathfrak{m}$ is a maximal, and therefore a prime, ideal. This is a contradiction. Therefore $I(PQ)= (1)$ if $I(P) = I(Q) = (1)$ . $\blacksquare$

Generalizations

One might be tempted to come to the more general conclusion that $I(PQ) = I(P) I(Q)$ , but this is false. For instance, if $a,b,c,d$ are indeterminates, then the ideal generated by the coefficients of $(ac)x^2 + (ad + bc)x + bd = (ax+b)(cx+d)$ is a proper subset of $(a,b)(c,d)$ .

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.

@@ Line 15: / Line 15: @@
 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>
-is not an element of <math>\mathfrak{m}</math>, a contradiction.  Therefore <math>I(PQ)= (1)</math> if <math>I(P) = I(Q) = (1)</math>.  <math>\blacksquare</math>
+is not an element of <math>\mathfrak{m}</math>, since <math>\mathfrak{m}</math> is a maximal, and therefore a prime, ideal.  This is a contradiction.  Therefore <math>I(PQ)= (1)</math> if <math>I(P) = I(Q) = (1)</math>.  <math>\blacksquare</math>
 == Generalizations ==
@@ Line 22: / Line 22: @@
 <cmath> (ac)x^2 + (ad + bc)x + bd = (ax+b)(cx+d) </cmath>
 is a proper subset of <math>(a,b)(c,d)</math>.
+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.
 [[Category:Ring theory]]
 [[Category:Commutative algebra]]
 [[Category:Number theory]]

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Difference between revisions of "Gauss's Lemma (polynomial)"

Revision as of 20:52, 21 September 2008

Proof

Generalizations