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Let $R$ be a commutative ring and $f\in R[X]$ a polynomial with $f\neq 0$ and suppose $a_1,...,a_n\in R$ are roots of $f$ with $a_i-a_j\in R^*$ for all $i,j$ with $1\leq i<j\leq n$.

How do I prove that $n\leq\deg f$?

I have no idea how to do this proof, maybe you can show me how?

user26857
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    Divide with remainder, $f(X) = (X - a_1)\cdot g(X) + r(X)$, with $\deg r < 1$. Then $f(a_1) = 0$ tells you what $r$ is. Then look whether you can see what the condition on the zeros tells you. – Daniel Fischer Sep 23 '15 at 17:16
  • @DanielFischer I don't really see what you did there –  Sep 23 '15 at 18:01
  • Do you know how to prove that a polynomial $f \in R[X]$ has at most $\deg f$ zeros in $R$ in the case that $R$ is a field? – Daniel Fischer Sep 23 '15 at 19:03
  • @DanielFischer No... –  Sep 23 '15 at 19:05
  • But $R$ isnt a field here right? –  Sep 23 '15 at 19:05
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    Right, $R$ is not a field here (it might be, that isn't forbidden, but we can't use anything specific to fields), so we need to adapt the proof, using the extra premises we have here. The answers to this question may be helpful to understand the proof for a field (more generally, for integral domains, the proof is the same). – Daniel Fischer Sep 23 '15 at 19:13
  • @DanielFischer would you mind posting an answer to show me a proof? –  Sep 23 '15 at 19:13
  • @user26857 $R^*$ is the group of all units in $R$, do you know what units are? –  Sep 23 '15 at 19:43
  • @user26857 An element $x$ in a ring is a unit if there exists an element $y\in R$ such that $xy=1$, where $1$ is the identity element of multiplication of $R$. –  Sep 23 '15 at 20:02

1 Answers1

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Induction on $n\ge 2$. The base case $n=2$ is left to the readers.

By the induction hypothesis $\deg f\ge n-1$. If $\deg f\ge n$ then there is nothing to prove, so we may assume $\deg f=n-1$. Write $f(X)=b_{n-1}X^{n-1}+\cdots+b_1X+b_0$ with $b_{n-1}\ne0$. We have $f(a_i)=0$ for $1\le i\le n$, that is, $b_{n-1}a_i^{n-1}+\cdots+b_1a_i+b_0=0$ for $1\le i\le n$. Set $$A=\pmatrix{1&1&\dots&1\\a_1&a_2&\dots&a_n\\\cdots&\cdots&\cdots&\cdots\\a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}}.$$ Then $(b_0\ b_1\ \dots\ b_{n-1})A=0$. But $\det A=\prod_{1\le i<j\le n}(a_j-a_i)\in R^*$, hence $A$ is invertible. Now we get $(b_0\ b_1\ \dots\ b_{n-1})=0$, a contradiction.

user26857
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  • Nicely written up. But you could have just as well started this induction at $n = 0$. (The condition for $a_i-a_j$ to be invertible is vacuous for $n \leq 1$, but there is nothing wrong about vacuous condition.) – darij grinberg Sep 28 '15 at 02:06