# Suppose that $f (X),g(X) = 1$. Prove that $f (X) = a_0$ for some $a_0$ with $a_0 \neq 0$

I am trying to prove the following:

Suppose that $f (X)g(X) = 1$ and further $f(X),g(X) \neq0$ . Prove that $f (X) = a_0$ for some $a_0$ with $a_0 \neq 0$

Very stuck on how to do this one.

Thoughts: My first thought was to multiply $f(X)g(X)$ but this is horrible and can't see where I would even go with it. I'm thinking maybe it has something to do with $f (X)g(X) = 1$ and there is something I'm not spotting!

Any hints?

• Are you assuming the coeff ring is a field or domain? Else it may fail, e.g. $\,(1\!-\!2x)(1\!+\!2x) \equiv 1\pmod 4.\ \$ Mar 26, 2015 at 23:04

Assuming that the the coefficients of the polynomials lie in a Field (or even in an Integral Domain), just apply the fact that the degree of $fg$ is the sum of the degrees of $f$ and $g$.

• Yes they lie in a field. So would this argument be sound. Since $f(X)g(X)=1$ and we have that $deg(f(X)g(X))=deg(f(X))+deg(g(X))$ then $0=deg(f(X))+deg(g(X))$ and since we define polynomials only with $deg(p(x)) \geq 0$ then we must have $deg(f(X))=0$ i.e. $f(X)$ is equal to some constant $a_0$ say. Also note $a_0 \neq 0$ since $f(X)g(X)=1$?
– Ryan
Mar 26, 2015 at 22:43

Write $f(X) = a_0 + a_1X +\cdots + a_nX^n$ and $g(X) = b_0 + b_1X + \cdots + b_mX^m$. Assume $n,m > 0$ with $a_n,b_m \neq 0$.

Then $1 = f(X)g(X) = a_0b_0 + (a_0b_1 + a_1b_0)X + \cdots + a_nb_mX^{n+m}$

Consequently $a_0b_0 = 1 \implies a_0 \neq 0$, and: $a_nb_m = 0$, contradiction.

Thus either $n = 0$, and we are done, or $m = 0$.

But if $g(X) = b_0$ then:

$1 = f(X)g(X) = a_0b_0 + a_1b_0X + \cdots a_nb_0X^n$ and we see:

$a_kb_0 = 0$ for $k = 1,2,\dots n$. Since $a_0b_0 = 1\neq 0$, we have $b_0 \neq 0$, so that:

$a_k = 0: \forall\ 1 \leq k \leq n$, so that $f(X) = a_0$.

(this is substantially ajotatxe's degree argument, without actually mentioning degree).