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Thread: 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 $\displaystyle \pm 1$ are roots , so you can divide it by $\displaystyle x^2-1$ , and again

    the quotient has $\displaystyle -1$ as a root so...etc.

    In short, $\displaystyle 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 $\displaystyle ab=1\Longrightarrow a^2b^2=(ab)^2=1$ 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, $\displaystyle 0=g(x)(x^5+x^4-x-1)$ for all $\displaystyle g(x)\in\mathbb{Q}[x]$. See which k can make g(x) really a polynomial on Q rather than a rational function?
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