module-finite extensions are integral

Theorem Suppose BA is module-finite. Then A is integral over B.

Proof. Choose uA.

For clarity, assume A is spanned by two elements ω1,ω2. The proof given clearly generalizes to the case where a spanning set for A has more than two elements.


uω1 =b11ω1+b12ω2
uω2 =b21ω1+b22ω2



and let Cadj be the adjugatePlanetmathPlanetmath of C. Then C(ω1ω2)=0, so CadjC(ω1ω2)=0.

Now, CadjC is a diagonal matrixMathworldPlanetmath with detC on the diagonal, so


where fB[x] is monic.

But neither ω1 nor ω2 is zero, so f(u) must be.

Note that, as with the field case, the converse is not true. For example, the algebraic integersMathworldPlanetmath are integral but not finite over .

Title module-finite extensions are integral
Canonical name ModulefiniteExtensionsAreIntegral
Date of creation 2013-03-22 17:01:23
Last modified on 2013-03-22 17:01:23
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 9
Author rm50 (10146)
Entry type Theorem
Classification msc 16D10
Classification msc 13C05
Classification msc 13B02
Related topic RingFiniteIntegralExtensionsAreModuleFinite