I don't understand how this proof works.

Question

My book proves "if $a(x)$ has degree $n$, it has at most $n$ roots." I will just copy the proof here.

Proof: If $a(x)$ had $n+1$ roots $c_1, \dots,c_{n+1}$, then by Theorem 2, $(x-c_1) \dots (x-c_{n+1})$ would be a factor of $a(x)$, and the degree of $a(x)$ would therefore be at least $n+1$. QED

Theorem 2 says if $a(x)$ has distinct roots $c_1, \dots ,c_m$ in $F$, then $(x-c_1) \dots (x-c_m)$ is a factor of $a(x)$, where $F$ is a field. All the polynomials are over the field $F$.

I just don't understand how that proof proves the theorem it tries to prove. Is it using PMI? Is there any other way to prove this?

Answer 1

This is because for any integral domain $A$ and any non-zero polynomials $f,g\in A[x]$, we have $$\deg f(x)g(x)=\deg f(x)+\deg g(x),$$ hence, if $p(x)$ divides $f(x)$, $\;\deg f(x)\ge\deg p(x)$.

In the present case, we would have $$\deg a(x)\ge\deg\bigl((x-c_1)\dotsm(x-c_{n+1}\bigr)=n+1.$$

Answer 2

Perhaps it's easier to consider a simple example. Suppose $f(x)=ax+b$ a degree one polynomial has two roots $c_1$ and $c_2$. Then $(x-c_1)(x-c_2)$ must divide $f(x)$. But $(x-c_1)(x-c_2)=x^2-(c_1+c_2)x+c_1c_2$ which has degree two. A degree two polynomial cannot divide a degree one polynomial. So $f$ cannot have two roots. The general proof is basically the same.

Answer 3

By the theorem if $\,f\ne 0\,$ has $\,n\!+\!1\,$ distinct roots $\,c_i\,$ then $\,f(x) = g(x) (x\!-\!c_1)\cdots (x\!-\!c_{n+1})\,$ for some $\,g\neq 0.\,$ Taking the degree of the RHS shows $\,\deg f \ge {n\!+\!1}.\,$ In particular $\,\deg f \neq n.$

Yes, there are other proofs, e.g. since $\,f\,$ is divisible by the nonassociate primes $\,x\!-\!c_i\,$ it must be divisible by their lcm = product, so $\,f\,$ has degree $\ge $ their product.

This property depends crucially on the coefficient ring being a domain (in fact it is equivalent to that). Else, e.g. $\,ab = 0\,$ for $\,a,b\ne 0,\,$ so $\, ax = 0\,$ has more roots $\,b,0\,$ than its degree $(= 1).$