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Thread: McCoy's Theorem Proof

  1. #1
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    McCoy's Theorem Proof

    Let R be a commutative ring with identity $\displaystyle 1 ( \neq 0 ) $ and let $\displaystyle f(x)= f_nx^n+ . . . + f_1x+f_0$ be a polynomial in $\displaystyle R[x]$

    Prove that $\displaystyle f(x)$ is a zero divisor of $\displaystyle R[x]$ if and only if there is a nonzero element $\displaystyle s \in R $ such that $\displaystyle sf(x) = (0) $

    Proof so far.

    Suppose that $\displaystyle f(x)$ is a zero divisor, then by definition $\displaystyle \exists g(x) \in R[x], g(x) \neq 0 $ such that $\displaystyle f(x)g(x) = 0$

    But how would I factor this down into one single element?

    Conversely, suppose that $\displaystyle sf(x)=(0) \ \ \ s \in R $, then would I be able to find a polynomial $\displaystyle g(x)$ that behaves like s?

    Thank you!
    Last edited by Jameson; Apr 30th 2009 at 10:40 AM. Reason: restored question
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let R be a commutative ring with identity $\displaystyle 1 ( \neq 0 ) $ and let $\displaystyle f(x)= f_nx^n+ . . . + f_1x+f_0$ be a polynomial in $\displaystyle R[x]$

    Prove that $\displaystyle f(x)$ is a zero divisor of $\displaystyle R[x]$ if and only if there is a nonzero element $\displaystyle s \in R $ such that $\displaystyle sf(x) = (0) $
    let $\displaystyle g(x)=a_mx^m + \cdots + a_0, \ a_m \neq 0.$ be a polynomial with minimum degree such that $\displaystyle f(x)g(x)=0.$ if $\displaystyle f_jg(x)=0, \ \forall j,$ then $\displaystyle f_ja_m=0, \ \forall j,$ and thus $\displaystyle a_mf(x)=0,$ and we're done.

    otherwise $\displaystyle \{j: \ f_jg(x) \neq 0 \} \neq \emptyset.$ let $\displaystyle \ell=\max \{j : \ f_jg(x) \neq 0 \}.$ then $\displaystyle 0=f(x)g(x)=(f_{\ell}x^{\ell} + \cdots + f_0)(a_mx^m + \cdots + a_0).$ thus $\displaystyle f_{\ell}a_m=0.$ so $\displaystyle f_{\ell}g(x)=f_{\ell}a_{m-1}x^{m-1} + \cdots + a_0.$

    hence $\displaystyle \deg f_{\ell}g(x) < m=\deg g.$ but we have: $\displaystyle f(x)(f_{\ell}g(x))=f_{\ell}f(x)g(x)=0,$ which is impossible because $\displaystyle g(x)$ was a polynomial with minimum degree satisfying $\displaystyle f(x)g(x)=0. \ \ \Box$
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  3. #3
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    Thank you!!!

    One last question:

    Prove that if R has no nonzero nilpotent elements and $\displaystyle f(x)g(x)=0$ for some $\displaystyle g(x)=g_mx^m+...+g_0 \in R[x] $, then $\displaystyle f_ig_j=0 \ \ \ \forall 0 \leq i \leq n,0 \leq j \leq n $.

    Proof so far.

    Suppose that for some element $\displaystyle r \in R$, if we have $\displaystyle r^k=0 $, then $\displaystyle r = 0 $.

    Now, let $\displaystyle f(x)g(x)=0$ for some $\displaystyle g(x)=g_mx^m+...+g_0 \in R[x] $

    So the product of each coefficient is also zero, how should I proceed?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post

    Prove that if R has no nonzero nilpotent elements and $\displaystyle f(x)g(x)=0$ for some $\displaystyle g(x)=g_mx^m+...+g_0 \in R[x] $, then $\displaystyle f_ig_j=0 \ \ \ \forall 0 \leq i \leq n,0 \leq j \leq \color{red}m $.
    proof by induction over $\displaystyle i+j.$ it's obvious from $\displaystyle f(x)g(x)=0$ that $\displaystyle f_0g_0=0.$ now let $\displaystyle 0 < \ell \leq m+n$ and suppose $\displaystyle f_rg_s=0,$ whenever $\displaystyle 0 \leq r+s < \ell.$ we need to show that $\displaystyle f_rg_s=0$ whenever

    $\displaystyle r+s=\ell.$ so suppose $\displaystyle r+s=\ell.$ the coefficient of $\displaystyle x^{\ell}$ in $\displaystyle f(x)g(x)$ is clearly $\displaystyle 0=\sum_{i < r, \ i+j=\ell}f_ig_j+ f_rg_s + \sum_{i > r, \ i+j=\ell} f_ig_j,$ which after mutiplying both sides by $\displaystyle f_rg_s$ gives us:

    $\displaystyle \sum_{i < r, \ i+j=\ell}f_rg_sf_ig_j+ (f_rg_s)^2 + \sum_{i > r, \ i+j=\ell}f_rg_sf_ig_j=0. \ \ \ \ \ \ \ \ \ \ \ (1)$

    now in the first sum in (1), since $\displaystyle i < r,$ we have $\displaystyle i+s < r+s=\ell.$ hence by induction hypothesis $\displaystyle f_ig_s=0.$ thus $\displaystyle f_rg_sf_ig_j=0,$ i.e. the first sum is equal to $\displaystyle 0.$ in the second sum in (1), since

    $\displaystyle i > r$ and $\displaystyle i+j=r+s=\ell,$ we must have $\displaystyle j < s.$ thus by induction hypothesis $\displaystyle f_rg_j=0,$ and hence $\displaystyle f_rg_sf_ig_j=0.$ so the second sum is $\displaystyle 0$ as well. hence (1) becomes: $\displaystyle (f_rg_s)^2=0,$ i.e. $\displaystyle f_rg_s$ is

    nilpotent. but $\displaystyle R$ has no nonzero nilpotent element. thus $\displaystyle f_rg_s=0. \ \ \Box$
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