product of finitely generated ideals

Let $R$ be a commutative ring having at least one regular elementand $T$ its total ring of fractions. Let $\\mathfrak{a}:=(a_{0},\\,a_{1},\\,\\ldots,\\,a_{m-1})$ and $\\mathfrak{b}:=(b_{0},\\,b_{1},\\,\\ldots,\\,b_{n-1})$ be two fractional ideals of $R$ (see the entry “fractional ideal of commutative ring”). Thenthe product submodule $\\mathfrak{ab}$ of $T$ is also a of $R$ and is generated by all the elements $a_{i}b_{j}$ , thus having a generating set of $m n$ elements.

Such a generating set may be condensed in the case of any Dedekind domain, especially for the of any algebraic number field one has the multiplication formula

\\displaystyle\\mathfrak{ab}=(a_{0}b_{0},\\,a_{0}b_{1}\\!+\\!a_{1}b_{0},\\,a_{0}b_{2%}\\!+\\!a_{1}b_{1}\\!+\\!a_{2}b_{0},\\,\\ldots,\\,a_{m-1}b_{n-1}).

(1)

Here, the number of generators is only $m\\!+\\!n\\!-\\!1$ (in principle, every ideal of a Dedekind domain has a generating system of two elements (http://planetmath.org/TwoGeneratorProperty)). The formula is characteristic (http://planetmath.org/Characterization) still for a wider class of rings $R$ which may contain zero divisors, viz. for the Prüfer rings (see [1]), but then at least one of $\\mathfrak{a}$ and $\\mathfrak{b}$ must be a regular ideal.

Note that the generators in (1) are formed similarly as the coefficients in the product of the polynomials $f(X):=f_{0}\\!+\\!f_{1}X\\!+\\cdots+\\!f_{m-1}X^{m-1}$ and $g(X):=g_{0}\\!+\\!g_{1}X\\!+\\cdots+\\!g_{n-1}X^{n-1}$ . Thus we may call the fractional ideals $\\mathfrak{a}$ and $\\mathfrak{b}$ of $R$ the coefficient modules $\\mathfrak{m}_{f}$ and $\\mathfrak{m}_{g}$ of the polynomials $f$ and $g$ (they are $R$ -modules). Hence the formula (1) may be rewritten as

\\displaystyle\\mathfrak{m}_{f}\\mathfrak{m}_{g}=\\mathfrak{m}_{fg}.

(2)

This formula says the same as Gauss’s lemma I for a unique factorization domain $R$ .

Arnold and Gilmer [2] have presented and proved the following generalisation of (2) which is valid under much less stringent assumptions than the ones requiring $R$ to be a Prüfer ring (initially: a Prüfer domain); the proof is somewhat simplified in [1].

Theorem (Dedekind–Mertens lemma). Let $R$ be a subring of a commutative ring $T$ . If $f$ and $g$ are two arbitrary polynomials in the polynomial ring $T[X]$ ,then there exists a non-negative integer $n$ such that the $R$ -submodules of $T$ generated by the coefficients of thepolynomials $f$ , $g$ and $f g$ satisfy the equality

\\displaystyle\\mathfrak{m}_{f}^{n+1}\\,\\mathfrak{m}_{g}=\\mathfrak{m}_{f}^{n}\\,%\\mathfrak{m}_{fg}.

(3)

References

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).