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Math Help - McCoy's Theorem Proof

  1. #1
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    McCoy's Theorem Proof

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

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

    Proof so far.

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

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

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

    Thank you!
    Last edited by Jameson; April 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 1 ( \neq 0 ) and let f(x)= f_nx^n+ . . . + f_1x+f_0 be a polynomial in R[x]

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

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

    hence \deg f_{\ell}g(x) < m=\deg g. but we have: f(x)(f_{\ell}g(x))=f_{\ell}f(x)g(x)=0, which is impossible because g(x) was a polynomial with minimum degree satisfying 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 f(x)g(x)=0 for some g(x)=g_mx^m+...+g_0 \in R[x] , then f_ig_j=0 \ \ \ \forall 0 \leq i \leq n,0 \leq j \leq n .

    Proof so far.

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

    Now, let f(x)g(x)=0 for some 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 f(x)g(x)=0 for some g(x)=g_mx^m+...+g_0 \in R[x] , then f_ig_j=0 \ \ \ \forall 0 \leq i \leq n,0 \leq j \leq \color{red}m .
    proof by induction over i+j. it's obvious from f(x)g(x)=0 that f_0g_0=0. now let 0 < \ell \leq m+n and suppose f_rg_s=0, whenever 0 \leq r+s < \ell. we need to show that f_rg_s=0 whenever

    r+s=\ell. so suppose r+s=\ell. the coefficient of x^{\ell} in f(x)g(x) is clearly 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 f_rg_s gives us:

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

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

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