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Let $K$ be an integral domain, and consider a polynomial $f$ in $K[x]$ with degree $\deg(f) \leq n$. Suppose there exist $n + 1$ pairwise distinct elements $\alpha_1, \ldots, \alpha_{n+1} \in K$ such that $f(\alpha_i) = 0$ for $i = 1, \ldots, n + 1$. Prove that $f$ must be the zero polynomial.

So the proof goes as follows:

Let's suppose $f\ne 0$. We have $f(\alpha_1)=0\Rightarrow f=(x-\alpha_1)q_1$ for some $q_1\in K[x]$. Let's take $f(\alpha_2)=(\alpha_2-\alpha_1)q_1(\alpha_2)$. As $K$ is an integral domein and $\alpha_2-\alpha_1\ne0$, then $q_1(\alpha_2)=0$. Then $q_1=(x-\alpha_2)q_2$ for some $q_2\in K[x]$. So $$f=(x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_{n+1})q_{n+1}, q_{n+1}\in K[x],$$ which contradicts $\deg(f)\le n$.

I can't really tell what's the specific property of integral domains that is needed for this result. Where are we using that $K$ is an integral domain (a commutative ring with identity which does not have divisors of zero)? We know that $\alpha_1,\dots,\alpha_{n+1}$ are pairwise distinct, so of course $\alpha_2-\alpha_1\ne0$. Why can't $K$ be just a ring?

Nij
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SAQ
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  • Okay now the related question HAS NOTHING TO DO with my question. I think I am being personally targeted right now. – SAQ Jan 23 '24 at 00:03
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    It uses the fact that nonzero elements are cancellable in a domain, i.e. $a_2\neq a_1\Rightarrow a_2-a_1\neq 0,$ so $(a_2-a_1)q(a_2)=0\Rightarrow q(a_2)=0.,$ This is highlighted in the proof of the Bifactor Theorem in the linked dupe (where a counterexample is given in non-domains). – Bill Dubuque Jan 23 '24 at 00:05
  • I found a counterexample. Let's $f(x)=x^2-\overline1\in\mathbb{Z}_8$. We have $f(\overline3)=\overline0, f(\overline1)=\overline0, f(\overline5)=\overline0,$ but the degree of $f$ is $2$. But I still don't understand where in this line of the proof: "As $K$ is an integral domein and $\alpha_2-\alpha_1\ne0$, then $q_1(\alpha_2)=0$" we're using that $K$ is a domain. – SAQ Jan 23 '24 at 00:18
  • That's the same non-domain counterexample that I linked in my prior comment. – Bill Dubuque Jan 23 '24 at 00:18
  • @BillDubuque, well I didn't see it directly from what you wrote. But what does cancellable in a domain mean? I really don't understand how we used it in the proof. – SAQ Jan 23 '24 at 00:19
  • @BillDubuque, we know that $a_1-a_2\ne0$ as they are pairwise distinct. – SAQ Jan 23 '24 at 00:20
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    The proof cancels a nonzero element, which is valid only because we are in a domain, i.e it uses $,ab=0,\ a\neq 0\Rightarrow b = 0,,$ for $,a = \alpha_2-\alpha_1,\ b = q(\alpha_2).\ \ $ – Bill Dubuque Jan 23 '24 at 00:24
  • @BillDubuque, thank you, I appreciate it. I understood it now. – SAQ Jan 23 '24 at 00:26
  • @BillDubuque, may I also ask you if $\mathbb{Z_8}$ is a commutative ring with unity? – SAQ Jan 23 '24 at 00:28
  • You might find helpful this constructive variant which shows how to split $n$ into nontrivial factors given a nonzero polynomial over $,\Bbb Z/n,$ with more roots than its degree. The Theorem there shows that being a domain is equivalent to this property of roots. – Bill Dubuque Jan 23 '24 at 00:29
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    @SAQ Just to add for example the polynomial $x^2-4$ in $\mathbb{Z}_{21}[x]$ has $2,13,5$ as roots – IrbidMath Jan 23 '24 at 00:30
  • Note $,a\neq 0,\ ab=0,\Rightarrow, b=0,$ is viewable as $\color{#c00}{{\rm cancelling}\ a},,$ because $,0 = a\cdot 0,,$ written explicitly we have that $,\color{#c00}a\color{#0a0}b = \color{#c00}a\cdot\color{#0a0} 0,\Rightarrow, \color{#0a0}{b=0}.\ $ Why do you doubt that $,\Bbb Z_8,$ Is commutative ring with $1?\ \ $ – Bill Dubuque Jan 23 '24 at 00:41
  • @IrbidMath No, the correct values are $\bmod 21!:, \sqrt{4}\equiv \pm 2,\pm5 \equiv 2,5,16,19,\ $ See here for how to compute them via CRT. – Bill Dubuque Jan 23 '24 at 00:46
  • @BillDubuque you are pushing yourself so hard am not trying to find all solution and I made a mistake. Take it easy. – IrbidMath Jan 23 '24 at 13:24

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