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It is known that a polynomial in an integral domain has at most as many roots as its degree. So conversely a polynomial with more roots that its degree has to be in a ring with zero divisors. For example in $\mathbb{Z}_8$ the equation $x^2 = 1$ has the solutions $1$, $3$, $5$ and $7$, and indeed $\mathbb{Z}_8$ has the zero divisors $2$, $4$ and $6$. Is there a way to construct these zero divisors from $1$, $3$, $5$ and $7$? More generally, given the roots $\{x_1, \dots, x_n\}$ of a polynomial $p \in R[X]$, how do I construct zero divisors of $R$ when deg $p < n$?

The most immediate idea is to write $p$ as $(X - x_1) \cdot \dots \cdot (X - x_{\textrm{deg } p})$ and then plugging in $x_n$ for $X$. Since $x_n$ is a root, this product equals $0$ and therefore consists of zero divisors. But writing $p$ as this product simply doesn't work in non-integral domains. If we take our example from before, then $(X - 1) \cdot (X - 3) = X^2 - 4X + 3$ and not $X^2 - 1$. But it feels like this idea should work, so how do I fix it?

LionCoder
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    your idea is very good actually. Remember that Ruffini still works, so given any root $p(\alpha)=0$ you can write $p(x)=(x-\alpha)q(x)$. If now $\beta$ is another root of $p$ but not of $q$, you get that $\beta-\alpha$ is a zero divisor – Exodd Jul 02 '22 at 19:59
  • @Exodd -- this should be an answer – HallaSurvivor Jul 02 '22 at 20:00
  • I assume Ruffini is using the Euclidean algorithm for polynomials and then showing that the remainder polynomial is zero, right? But why then can I use the Euclidean algorithm, if my ring is not an integral domain? – LionCoder Jul 02 '22 at 20:04
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    Because the polynomial you're dividing by is monic, so that one never needs to divide to compute the next term. – Greg Martin Jul 02 '22 at 20:09
  • You're right! Good point. But now, why does there exist a $\beta$ that is a root of $p$ and not of $q$? – LionCoder Jul 02 '22 at 20:21
  • because you can repeat Ruffini until you get a total decomposition of $p$ with less roots than you have OR a zero divisor directly. – Exodd Jul 02 '22 at 20:47
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    Explaining a bit more formally, suppose you have $p(x) = (x-\alpha_1)\dots(x-\alpha_k)q_k(x)$ and another root $\beta$. If $q_k(\beta) = 0$, then you can still decompose it with $(x-\beta)$, otherwise $q_k(\beta)$ is necessarily a divisor of zero. If you continue factorizing, eventually $q(x) = 1$ and you get the divisor nonetheless – Exodd Jul 02 '22 at 20:50
  • See the proof in this answer in the linked dupe. – Bill Dubuque Jul 03 '22 at 06:17

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