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A polynomial $f$ of degree $n$ over a field $F$ has at most $n$ roots in $F$.

My attempt: Let $f\in F[x]\setminus \{0\}$ and $\deg (f)=n$. Assume towards contradiction, $\exists \{c_1,…,c_m\}\subseteq F$ ($n\lt m$) such that $f(c_i)=0_F$, $\forall i\in J_m$. In particular, $f(c_1)=0_F$. By theorem 4 corollary 1 section 4.4, $\exists q_1\in f[x]$ such that $f=(x-c_1)\cdot q_1$. Since $f(c_2)=0_F$, we have $f(c_2)=(c_2-c_1)\cdot q_1(c_2)=0_F$, by theorem 2 section 4.2. Which implies $c_2-c_1=0_F$ or $q_1(c_2)=0_F$. Since $c_1\neq c_2$, we have $q_1(c_2)=0_F$. By theorem 4 corollary 1 section 4.4, $\exists q_2\in F[x]$ such that $q_1=(x-c_2)\cdot q_2$. By mathematical induction, $q_i=(x-c_{i+1})\cdot q_{i+1}$, $\forall 1\leq i\leq m-1$. So $f=(x-c_1)\dotsb (x-c_m)\cdot q_{m+1}$. Thus $n=\deg (f)=\deg [(x-c_1)\dotsb (x-c_m)\cdot q_{m+1}]=\deg (x-c_1)+…+\deg (x-c_m)+\deg (q_{m+1})=m+\deg (q_{m+1})$. Which implies $n-m=\deg (q_{m+1})\geq 0$, i.e. $n\geq m$. Which contradicts our initial assumption of $n\lt m$. Hence $\exists$ at most $n$ roots in $F$. Is my proof correct?

user264745
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    Are we all supposed to now the contents of theorem 4 corollary 1 section 4.4? – José Carlos Santos Nov 30 '22 at 14:13
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    @JoséCarlosSantos it’s a elementary/standard result. I state it for reference. – user264745 Nov 30 '22 at 14:22
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    Your proof may be considered as correct, but better make clear at the beginning that $c_1,\dots,c_m$ are distinct, and at the end that $q_{m+1}\ne0$ (for its degree to be $\ge0$). – Anne Bauval Nov 30 '22 at 14:52
  • @AnneBauval yess that’s a nice point. First, I thought to write $c_i\neq c_j$, if $i\neq j$, latter I changed my mind and wrote ${c_1,…,c_m}$, since by definition elements of a set are different. I could have included details of $q_i\neq 0$, $\forall 1\leq i\leq m+1$. But that’s trivial because if $q_j=0$, then $f=0$. Thus we reach contradiction. – user264745 Nov 30 '22 at 15:30
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    I know the details you could/should have included. And I don't think that "by definition elements of a set are different". – Anne Bauval Nov 30 '22 at 15:32
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    you may want to try proving this by relating $f$ to a characteristic polynomial for an $n\times n$ matrix and then using the fact that eigenvectors for different eigenvalues are linearly independent – user8675309 Nov 30 '22 at 16:35
  • @user8675309 I think characteristic polynomial of $f$ is of degree one. Can you write matrix representation of $f$? It is $1\times 1$ matrix. – user264745 Nov 30 '22 at 18:21
  • Let me try again-- you can assume that $f$ is a monic polynomial of degree $n$ (i.e. divide out leading coefficient if needed, this doesn't change roots). Now let $f$ be the characteristic polynomial for some $n\times n$ matrix (e.g. I like the Companion Matrix). Now think through the linearly independent eigenvectors argument... – user8675309 Nov 30 '22 at 18:50
  • @user8675309 I'm sorry. I can't fill in details. Can you elaborate more or write proof? – user264745 Nov 30 '22 at 22:00

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