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I am trying to prove the following statement:

If the polynomial $f(x)$ of degree $n$ has roots $a_1,a_2,...,a_k$ with multiplicities $\alpha_1,\alpha_2,...,\alpha_k$ in a field $F$ (with $\alpha_i\geq 1$), then $f(x)$ has $(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}$ as a factor. In particular, a polynomial of degree $n$ in one variable over a field $F$ has at most $n$ roots in $F$, even counted with multiplicity.

This is the proof that I wrote

For the first part, we proceed by induction on $k$. If $k=1$, then by definition, we may write $f(x)=(x-a_1)^{\alpha_1}q(x)$. Now, suppose that the statement is true for up to $k$ roots, and consider a collection of $k+1$ roots. Using the inductive hypothesis, we may write $$f(x)=(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}q(x)\quad\text{ and }\quad f(x)=(x-a_{k+1})^{\alpha_{k+1}}q'(x)$$ Equating $$(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}q(x)=(x-a_{k+1})^{\alpha_{k+1}}q'(x)$$ and using the fact that $(x-a_{k+1})$ is an irreducible element that is distinct from any on the left hand side gives us that $(x-a_{k+1})^{\alpha_{k+1}}\mid q(x)$, which proves the first statement.

Now, suppose that a polynomial $f(x)$ of degree $n$ as more than $n$ roots in $F$, counted with multiplicity. From what we showed, we must have $(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}\mid f(x)$. But $F[x]$ is an integral domain so this is impossible, as the divisor has a higher degree than $f(x)$.

I was first wondering if the proof was correct. Moreover, in proving the first half of the statement I used the fact that an irreducible element is prime, which leverages the fact that $F[x]$ is a U.F.D. Is there a more general method of proof that works on general integral domains?

Vasting
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  • Remember your basic polynomial division. Let $a$ be arbitrary. Then $p(x)=(x-a)q(x)+r$ where $r$ is a constant (polynomial division). Then $p(a)=r$ and $a$ is a root iff $r=0$. You can use this in various contexts as the basis for an induction on degree. – Mark Bennet May 18 '20 at 21:06
  • @MarkBennet I tried something like that but ran into issues when dealing with multiplicities; I'll take another look. Also, is what I wrote correct? – Vasting May 18 '20 at 21:13
  • What is your definition of a "root of multiplicity $n$?" – Bill Dubuque May 18 '20 at 21:20
  • @Gone The one I was given is that $a$ is a root of multiplicity $m$ if $a$ is a root and $f(x)$ is divisible by $(x-a)^m$ but not by $(x-a)^{m+1}$. – Vasting May 18 '20 at 21:21
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    Then your proof is correct, but you should give more details, e.g. you seem to be implicitly using unique factorization after "is an irreducible..." Here you should show that the $x-a_i$ are nonassociate primes since the $a_i$ are distinct. See also the BiFactor Theorem. – Bill Dubuque May 18 '20 at 21:31
  • Essentially you are using a special case of: in a UFD: lcm = product for pairwise coprimes (here the coprimes are powers of nonassociate primes $,x−a_1),,$ i.e. the immediate extension of this proof to account for multiplicity. – Bill Dubuque May 18 '20 at 21:55
  • @Gone In the proof you linked, where you show that $D$ is a domain $\iff$ every polynomial $f(x)\neq 0\in D[x]$ has at most $\deg f$ roots, are those roots all distinct or is multiplicity accounted? The proof you mentioned iteratively applies the factor theorem, which I am not sure how to do when roots have $>1$ multiplicity. – Vasting May 18 '20 at 21:59
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    Those proofs don't handle multiplicity, but we could trivially extend them to do so if we employ a recursive definition of multiplicity such as: $r$ is a root of $f$ of mult. $n>1$ if $,f(r) = 0,$ and $r$ is a root of mult. $n-1$ of $,f/(x!-!r)\ \ $ – Bill Dubuque May 18 '20 at 22:11
  • It does depend a bit on what is meant by multiplicity and how it might be detected or known. But $f(x)=(x-a)^mq(x)+r(x)$ where $r(x)$ has degree less than $m$. Now $f(a)=0$ so that $r(a)=0$ and $r(x)$ has a factor $(x-a)$ which can be extracted, and this can be repeated. – Mark Bennet May 19 '20 at 05:44

1 Answers1

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I'd state the result as follows:

If $F$ is a field and $f \in F[x]$ not zero, then $f(x)=(x-a_1)\cdots(x-a_m)g(x)$, where $m\ge 0$, $a_i \in F$, and $g$ has no roots in $F$. In particular, $m \le n$, the degree of $f$.

and prove it by induction on the degree of $f$:

If $f$ has no roots in $F$, then take $m=0$ and $g=f$. In particular, this holds for $f$ nonzero constant.

If $f$ has a root $a_1$ in $F$, then $f(x)=(x-a_1)q(x)$ by polynomial division. The result follows by induction applied to $q$, whose degree is less than that of $f$.

No need to bring multiplicities into this argument until now. But then it is easy to conclude a result about multiplicities by grouping equal factors $x-a_i$.

Note that unique factorization is not a part of the argument.

lhf
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  • The heart of the proof that the OP seeks lies within your "it is easy to conclude". It it was so "easy" the OP would not have asked the question. – Bill Dubuque May 19 '20 at 00:26