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- Homework Statement:
- Let ##R## be a commutative ring with unity. Let ##p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0## be an element of the polynomial ring ##R[x]##. Prove that ##p(x)## is a zero divisor in ##R[x]## if and only if there is a nonzero element ##b \in R## such that ##bp(x) = 0##. Hint: Let ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## be a nonzero polynomial of minimal degree such that ##g(x)p(x) = 0##. Show that ##b_ma_n = 0## and so ##a_ng(x)## is a polynomial of degree less than ##m## that also gives ##0## when multiplied by ##p(x)##. Conclude that ##a_ng(x) = 0##. Apply a similar argument to show by induction on ##i## that ##a_{n-i}g(x) = 0## for ##i = 0, 1, \dots, n## and show that this implies ##b_mp(x) = 0##.

- Relevant Equations:
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Proof: ##(\Leftarrow)## Suppose there exists non zero ##b \in R## such that ##bp(x) = 0##. Well, ##R \subset R[x]##, and so by definition of zero divisor, ##p(x)## is a zero divisor. (assuming ##p(x) \neq 0##).

##(\Rightarrow)## Suppose ##p(x)## is a zero divisor in ##R[x]##. Then we can choose a nonzero polynomial of minimal degree ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## such that ##g(x)p(x) = 0##. From the reading, the coefficient of ##x^{m+n}## of ##g(x)p(x)## is

$$ \sum_{k = 0}^{n+m} a_kb_{n+m-k} = a_nb_m$$

since for ##k > n## we have ##a_k = 0## and for ##k < n## we have ##b_{n+m-k} = 0##. Since ##g(x)p(x) = 0## we can conclude ##a_nb_m = 0##, I think? Now, ##\deg a_ng(x) < \deg g(x)## and ##(a_ng(x))p(x) = a_n(g(x)p(x)) = a_n\cdot 0 = 0##. This implies ##a_ng(x) = 0## since it has degree less than ##g(x)## and thus is not a zero divisor of ##p(x)##.

We will show in general that for all ##i = 0, 1, \dots, n##, ##a_{n-i}g(x) = 0##. We've shown this for ##i = 0## and proceed by induction. I am stuck here. Any advice, please?

Also a question, is ##R[[x]] \subseteq R[x]## where ##R[[x]]## is the ring of formal power series with coefficients from ##R##? By the wikipedia, it seems every polynomial in ##R[x]## has finite degree, but I just wanted to confirm.

##(\Rightarrow)## Suppose ##p(x)## is a zero divisor in ##R[x]##. Then we can choose a nonzero polynomial of minimal degree ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## such that ##g(x)p(x) = 0##. From the reading, the coefficient of ##x^{m+n}## of ##g(x)p(x)## is

$$ \sum_{k = 0}^{n+m} a_kb_{n+m-k} = a_nb_m$$

since for ##k > n## we have ##a_k = 0## and for ##k < n## we have ##b_{n+m-k} = 0##. Since ##g(x)p(x) = 0## we can conclude ##a_nb_m = 0##, I think? Now, ##\deg a_ng(x) < \deg g(x)## and ##(a_ng(x))p(x) = a_n(g(x)p(x)) = a_n\cdot 0 = 0##. This implies ##a_ng(x) = 0## since it has degree less than ##g(x)## and thus is not a zero divisor of ##p(x)##.

We will show in general that for all ##i = 0, 1, \dots, n##, ##a_{n-i}g(x) = 0##. We've shown this for ##i = 0## and proceed by induction. I am stuck here. Any advice, please?

Also a question, is ##R[[x]] \subseteq R[x]## where ##R[[x]]## is the ring of formal power series with coefficients from ##R##? By the wikipedia, it seems every polynomial in ##R[x]## has finite degree, but I just wanted to confirm.