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Math Help - Invariant factor decomposition from annihilators

  1. #1
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    Invariant factor decomposition from annihilators

    I want to know how to find the invariant factor decompostion of
    M \cong \frac{R}{((x-1)^3)}\oplus \frac{R}{((x^2+1)^2)} \oplus \frac{R}{((x-1)(x^2+1)^4)} \oplus\frac{R}{((x+2)(x^2+1)^2)}

    If it is relevant, in a previous part of the question I found the primary decomposition to be
    <br />
\frac{R}{((x-1)^3)} \oplus \frac{R}{((x^2+1)^2)} \oplus \frac{R}{(x-1)} \oplus \frac{R}{((x^2+1)^4)} \oplus \frac{R}{(x+2)} \oplus \frac{R}{((x^2+1)^2)}

    edit: forgot to mention that R = \mathbb{R}[x]
    Last edited by badgerigar; June 15th 2009 at 05:27 PM. Reason: forgot R
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  2. #2
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    Quote Originally Posted by badgerigar View Post
    I want to know how to find the invariant factor decompostion of
    M \cong \frac{R}{((x-1)^3)}\oplus \frac{R}{((x^2+1)^2)} \oplus \frac{R}{((x-1)(x^2+1)^4)} \oplus\frac{R}{((x+2)(x^2+1)^2)}

    If it is relevant, in a previous part of the question I found the primary decomposition to be
    <br />
\frac{R}{((x-1)^3)} \oplus \frac{R}{((x^2+1)^2)} \oplus \frac{R}{(x-1)} \oplus \frac{R}{((x^2+1)^4)} \oplus \frac{R}{(x+2)} \oplus \frac{R}{((x^2+1)^2)}

    edit: forgot to mention that R = \mathbb{R}[x]
    M \cong \frac{R}{<(x^2+1)^2>} \oplus \frac{R}{<(x-1)(x^2+1)^2>} \oplus \frac{R}{<(x-1)^3(x+2)(x^2+1)^4>}.
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  3. #3
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    Ok, so I'm guessing that \frac{R}{<(x-1)^3(x+2)(x^2+1)^4>} is there because the denominator is the lowest common multiple of the original denominators. But how did you find the other 2 terms? They appear to have the same product as the first set of denominators, is that relevant?
    Last edited by badgerigar; June 16th 2009 at 07:52 PM. Reason: added math tags
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  4. #4
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    I was going through my notes for a different question and found a worked example. On the off chance that somebody else was wondering about this also, the method is to take the highest power for each term and multiply them and then the next highest power for each term and so on until you run out of terms. So in this case we have:



    Taking the highest power of each term gives

    \frac{R}{<(x-1)^3(x+2)(x^2+1)^4>}

    and leaves us with
    \frac{R}{((x^2+1)^2)} \oplus \frac{R}{(x-1)} \oplus \frac{R}{((x^2+1)^2)}

    Taking the highest powers remaining gives
    \frac{R}{<(x-1)(x^2+1)^2>}

    and leaves
    <br />
\frac{R}{<(x^2+1)^2>}

    So the invariant factor decomposition is
    \frac{R}{<(x^2+1)^2>} \oplus \frac{R}{<(x-1)(x^2+1)^2>} \oplus \frac{R}{<(x-1)^3(x+2)(x^2+1)^4>}
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