Compute the invariant factors and elementary divisors of a given module

Suppose the vector space V is the direct sum of cyclic F[x]-modules whose annihilators are (x+1)^2, (x-1)(x^2+1)^2, x^4-1, and (x+1)(x^2-1). Determine the invariant factors and elementary divisors of V.


We have V \cong F[x]/((x+1)^2) \oplus F[x]/((x-1)(x^2+1)^2) \oplus F[x]/(x^4-1) \oplus F[x]/((x+1)(x^2-1)). If x^2+1 is irreducible over F, then by the Chinese Remainder Theorem for modules (see here and here), we have V \cong F[x]/((x+1)^2) \oplus F[x]/(x-1) \oplus F[x]/((x^2+1)^2) \oplus F[x]/(x+1) \oplus F[x]/(x-1) \oplus F[x]/(x^2+1) \oplus F[x]/((x+1)^2) \oplus F[x]/(x-1).

The elementary divisors of V are thus (x^2+1)^2, x^2+1, (x+1)^2, (x+1)^2, x+1, x-1, x-1, and x-1.

The invariant factors of V are (x^2+1)^2(x+1)^2(x-1), (x^2+1)(x+1)^2(x-1), and (x+1)(x-1).

If x^2+1 is reducible over F, then is has a root \alpha, and indeed x^2+1 = (x+\alpha)(x-\alpha). Now by the Chinese Remainder Theorem for modules we have V \cong F[x]/((x+1)^2) \oplus F[x]/(x-1) \oplus F[x]/((x+\alpha)^2) \oplus F[x]/((x-\alpha)^2) \oplus F[x]/(x+1) \oplus F[x]/(x-1) \oplus F[x]/(x+\alpha) \oplus F[x]/(x-\alpha) \oplus F[x]/((x+1)^2) \oplus F[x]/(x-1).

So the elementary divisors of V are (x-\alpha)^2, x-\alpha, (x+\alpha)^2, x+\alpha, (x+1)^2, (x+1)^2, x+1, x-1, x-1, and x-1.

The invariant factors of V are (x-\alpha)^2(x+\alpha)^2(x+1)^2(x-1), (x-\alpha)(x+\alpha)(x+1)^2(x-1), and (x+1)(x-1). (Which are the same as if F did not contain \alpha. This is to be expected in light of Corollary 18 on page 477 of D&F.)

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Comments

  • Marko Milosevic  On December 14, 2011 at 8:09 am

    You need to consider characteristic 2, then $( x+1)$ factors to $(x+i)^2$

    • nbloomf  On December 14, 2011 at 4:49 pm

      All that matters here is how the polynomials factor over F.

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