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Where I've got to so far:

Let $m(X)$ ε I be a non-zero polynomial of least degree.

For any $a(X)$ ε $\mathbb{F}$$[X]$, $a(X)m(X)$ ε I as I is a principal ideal of $\mathbb{F}$$[X]$ thus $<m(x)>$ $⊆$ I.

Let us now consider $b(X)$ ε I. By the division theorem for polynomials, $b(X) =q(x)m(x) + r(X)$ where $q(X), r(X)$ ε $\mathbb{F}$$[X]$ and $r(X) = 0$ or $deg(r(X)) < deg(m(X))$.

Observe that $r(X) = b(X) - q(X)m(X)$ ε I.

If $r(X) = 0$ then we are done: We have shown that $b(X) = q(X)m(X)$ which means $I ⊆ <m(X)>$ hence I $=$ $<m(X)>$.

If $deg(r(X)) < deg(m(X))$ then observe that $m(X)$ was chosen to have the least degree in I thus $deg(r(X)) = 0$ which means $r(X)$ is a constant polynomial.

So where I am so far is $r(X)$ = $b(X) - q(X)m(X)$ where $r(X)$ is a constant polynomial. How do deduce $b(X) = q(X)m(X)$ in this situation?

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  • $\begingroup$ Hint: Consider a non-zero polynomial in $I$ of least degree, and proceed by induction on the degree of polynomials in $I$. $\endgroup$
    – Bernard
    Oct 30, 2019 at 17:33
  • $\begingroup$ @Bernard What does least degree mean? $\endgroup$
    – SFR
    Oct 30, 2019 at 17:34
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    $\begingroup$ The set of degrees of the polynomials in $I$ is a subset of $\mathbf N$, and as such, it has a smallest element. This means that you consider a polynomial of degree, say $d$, such that all other polynomials in $I$ have degree $\ge d$. $\endgroup$
    – Bernard
    Oct 30, 2019 at 17:38
  • $\begingroup$ You should take a look at this before anything else. $\endgroup$
    – rschwieb
    Oct 30, 2019 at 18:12
  • $\begingroup$ @Bernard, could you help me out please? I'm stuck. :/^ $\endgroup$
    – SFR
    Oct 31, 2019 at 13:17

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