Showing $(q(X),q'(X)) = F[X]$

Question

I want to show that supposing $F$ is a field, and $q'(X)$ is the derivative of the polynomial $q(X)$ and if $q(X) \in F[X]$ is irreducible, then $(q(X),q'(X)) = F[X]$ where $(q(X),q'(X))$ is an ideal. I can see that clearly $(q(X),q'(X)) \subseteq F[X]$ but I have troubles seeing the other inclusion. Any suggestions or hints are greatly appreciated. Thanks!

Answer 1

We assume in the following that $\deg q(X) \ge 2$; if $\deg q(X) < 2$, then either $\deg q(X) = 1$, in which case $\deg q'(X) = 0$ and $q'(X) \ne 0$ is a unit in $F$, or $\deg q(X) = 0$ and $0 \ne q(X) \in F$ is a unit; in either case we trivially have $\langle q(X), q'(X) \rangle = F[X]$.

$F$ being a field, we know that $F[X]$ is a principle ideal domain.

Since $F[X]$ is a PID, the ideal

$\langle q(X), q'(X) \rangle = \langle d(X) \rangle \tag 1$

for some $d(X) \in F[X]$.

We will show that $d(X)$ is a unit in $F[X]$. Suppose, the the contrary, $d(X)$ is not a unit. We have

$d(X) \mid q(X) \; \text{and} \; d(X) \mid q'(X); \tag 2$

thus, since $q(X)$ is irreducible in $F(X)$, and $d(X)$ is not a unit,

$q(X) = ud(X), \tag 3$

for some unit $u \in F(X)$; it follows that

$\deg d(X) = \deg q(X); \tag 4$

noting that

$\deg q'(X) = \deg q(X) -1, \tag 5$

we see via (4) that

$d(X) \not \mid q'(X); \tag 6$

this contradicts (1), (2) and thus our assertion that $d(X)$, the generator of the principle ideal $\langle d(X) \rangle = \langle q(X), q'(X) \rangle$, is not a unit is false; thus there is $v \in F$ with

$vd(X) = 1, \tag 7$

so

$ \langle q(X), q'(X) \rangle = \langle d(X) \rangle = \langle 1 \rangle = F[X], \tag 7$

as was to be demonstrated.

Answer 2

It follows from the fact that $K(X)$ is an Euclidean domain. Using the general Euclidean division, it can be seen that there exist two polynomials $a(X)$ and $b(X)$ such that $a(X)q(X)+b(X)q'(X)=\text{GCD}(q,q')$ (where in this case a polynomial is a greatest common divisor of two polynomials $c(X)$ and $d(X)$ if it divides both and has the largest degree). Since you assumed $q$ to be irreducible, $\text{GCD}(q,q')=1$.