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Math Help - Zero divisors and such

  1. #1
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    Zero divisors and such

    We have elements a in the ring R=Q[x]/(x^5+x^4-x-1)

    If a is nilpotent, find the smallest positive k with a^k=0
    If a is not nilpotent, but a zero divisor, find a non-zero be that is smallest so that ab=0
    If it is not a zero divisor, find its inverse.

    i) a=x^3+x^2+x+1
    ii) a=x
    iii) a=x^2
    iv) a=x^4-1

    I don't think any are nilpotent...I've tried various ks to no avail, but I could be wrong
    i) I got a zero divisor with b=x^2-1
    I think ii) has an inverse since x^5+x^4-x-1=0, then x^5+x^4-x=1, so that x(x^4+x^3-1)=1, and so it has inverse x^4+x^3-1.
    iii) I'm not sure here
    iv) I got a zero divisor with b=x+1

    Do these look right, or am I missing someway any of these are nilpotents, and what about iii)? Thank you.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    We have elements a in the ring R=Q[x]/(x^5+x^4-x-1)

    If a is nilpotent, find the smallest positive k with a^k=0
    If a is not nilpotent, but a zero divisor, find a non-zero be that is smallest so that ab=0
    If it is not a zero divisor, find its inverse.

    i) a=x^3+x^2+x+1
    ii) a=x
    iii) a=x^2
    iv) a=x^4-1

    I don't think any are nilpotent...I've tried various ks to no avail, but I could be wrong
    i) I got a zero divisor with b=x^2-1
    I think ii) has an inverse since x^5+x^4-x-1=0, then x^5+x^4-x=1, so that x(x^4+x^3-1)=1, and so it has inverse x^4+x^3-1.
    iii) I'm not sure here
    iv) I got a zero divisor with b=x+1

    Do these look right, or am I missing someway any of these are nilpotents, and what about iii)? Thank you.

    Why don't you first try to decompose the polynomial as much as you can? It's obvious that \pm 1 are roots , so you can divide it by x^2-1 , and again

    the quotient has -1 as a root so...etc.

    In short, x^5+x^4-x-1=(x+1)^2(x-1)(x^2+1)\Longrightarrow (iv) is nilpotent (why?), (i) is a zero divisor (you'll see this when you decomposed as explained above) and (ii)-(iii) are units (if ab=1\Longrightarrow a^2b^2=(ab)^2=1 since there's commutativity here.)

    BTW, to ask for "smallest b s.t....etc." is nonsense here. Perhaps you meant "polynomial b of smallest degree..."?

    Tonio
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  3. #3
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    Okay, I think I get it.

    I am just a little confused about parts of it...

    I got i, ii....for iii, you are saying that the inverse of x^2 would have to be (x^4+x^3-1)^2? Also, I do not see what k would have to be to get a^k=0.
    I see how it could be a zero divisor by long division, but not sure about this.

    Thank you.
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  4. #4
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    a*b=1, then (a*b)*(a*b)=1. Multiplication is commutative here, so a^2*b^2=1. There's no reason that b^2 can't be the multiplicative inverse of a^2.
    To compute the desired k, remember, in ring R, 0=g(x)(x^5+x^4-x-1) for all g(x)\in\mathbb{Q}[x]. See which k can make g(x) really a polynomial on Q rather than a rational function?
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