1. ## 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!

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 $f(x)g(x)=0. \ \ \Box$

3. 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?

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 $f_rg_s=0. \ \ \Box$

### mc coy theorem

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