Tuesday 5 September 2017

abstract algebra - Prove that if $f,gin F[x]$ are non-constant polynomials, then $f(g)$ is a non-constant polynomial




For any polynomials $f$, $g$ $\in F[x]$ (where $F$ is a field), let $f(g)$ denote the polynomial obtained by substituting $g$ for the variable $x$ in $f$.
t,
Suppose $f,g \in F[x]$ are non-constant polynomials (by non-constant, I mean they are of degree $> 0$. So, for example, $f(x) = 5$ is constant, while $g(x) = x^4 + 2x$ is not). I need to prove that $f(g)$ is also a non-constant polynomial.



It seems like an extremely intuitive thing, and my thoughts are that it should be shown using a degree argument.



I.e., if $f,g$ are non-constant polynomials, then $deg(f)\geq 1$ and $deg(g) \geq 1$. So, I was thinking that I could show that then $deg(f(g))\geq 1$ as well, but I am not sure what the actual mechanics of showing this should be.



Could anybody please let me know how to go about proving this? Thank you ahead of time for your time and patience!


Answer




Show that if $f$ is of degree $n$ and if $g$ is of degree $m$, then $f(g)$ is of degree $mn$. To this end, note that if $f = \lambda X^n + \cdots$ and if $g = \mu X^m+\cdots$ then the unique term of degree $mn$ in $f(g)$ is $\lambda \mu^m X^{mn}$, and this is nonzero since $\lambda\mu^m\neq 0$.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...