proof of Dedekind-Mertens lemma

Let R be subring of the commutative ring T and


be arbitrary polynomialsPlanetmathPlanetmath in T[X].  We will prove by induction on n that the R-submodulesMathworldPlanetmath of T generated by the coefficients of the polynomials f, g, and fg satisfy

Mfn+1Mg=MfnMfg (1)

where the product modules are generated by the products of their generatorsPlanetmathPlanetmath.

The generators of the right hand side of (1) belong obviously to the left hand side, whence only the containment

Mfn+1MgMfnMfg (2)

has to be proved.

Firstly, (2) is trivial in the case  n=0.  Let now  n>0.  Define

fj:= 0forj<0orj>m

and let Gn be the R-submodule of T generated by g0,g1,,gn-1.  We have


where hk is the coefficient of Xk of the polynomial fg, and thus by induction we can write


This implies the containment


for every i.  In addition, we have




From this we infer that


is true for each  i = m, m-1,, 0.  Thus also (2) is true.


  • 1 J. Pahikkala: “Some formulae for multiplying and inverting ideals”.  – Ann. Univ. Turkuensis 183 (A) (1982).
  • 2 J. Arnold & R. Gilmer: “On the contents of polynomials”.  – Proc. Amer. Math. Soc. 24 (1970).
  • 3 T. Coquand: “A direct proof of Dedekind–Mertens lemma”. University of Gothenburg 2006. (Available coquand/mertens.pdfhere.)
Title proof of Dedekind-Mertens lemma
Canonical name ProofOfDedekindMertensLemma
Date of creation 2013-12-15 20:29:20
Last modified on 2013-12-15 20:29:20
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 5
Author pahio (2872)
Entry type Proof
Classification msc 13A15
Classification msc 13M10
Classification msc 16D10
Classification msc 16D25