Polynomial with degree less than degree of an irreducible polynomial of the same root is 0

Question

Let $F$ be a field, and $p(x)\in F[x]$ be an irreducible polynomial. Suppose $\alpha$ is a root of $p(x)$. Show that if $q(x)$ is a polynomial such that $\deg q(x) < \deg p(x)$ and $q(\alpha)=0$, then $q(x)=0$.

I tried to use the Division Algorithm, assuming $q(x)\neq 0$,

$p(x) = f(x) q(x) + r(x)$, where $\deg r(x) < \deg q(x)$.

Then I substituted $x=\alpha$, but it did not lead to anything meaningful. How can I use the fact that $p(x)$ is irreducible.

Thank you.

Answer 1

You're one the right track, but you need to follow through!

Substituting $x=\alpha$ and using the fact that $p(\alpha)=q(\alpha)=0$, you can conclude that $r(\alpha) = 0$. Thus, you have found a polynomial of strictly smaller degree that also claims $\alpha$ as a root. Now, apply the division algorithm to $q(x)$ and $r(x)$ as in the Euclidean algorithm. Continuing like this, you will generate a sequence of polynomials $$ r_0(x) = q(x), \quad r_1(x) = r(x), \quad r_2(x), \quad \dots $$ of strictly decreasing degree, each of which has $\alpha$ as a root. For some $t$, you arrive at $r_t(x)=0$. (Do you see why?)

Once $\deg r_s \le 0$ for some $s$, since $\alpha$ is a root and $r_s$ is a constant, it must be that $r_s = 0$.

Then, $d(x) = r_{t-1}(x)$ is a common divisor of all $r_s(x)$. This contradicts $p(x)$ being irreducible.