# A question (given as an image) on how to show a polynomial ring is equal to a principal ideal.

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 $$$$ $$⊆$$ 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 $$I ⊆ $$ hence I $$=$$ $$$$.

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?

• Hint: Consider a non-zero polynomial in $I$ of least degree, and proceed by induction on the degree of polynomials in $I$. Oct 30, 2019 at 17:33
• @Bernard What does least degree mean?
– SFR
Oct 30, 2019 at 17:34
• 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 $\ge d$. Oct 30, 2019 at 17:38
• You should take a look at this before anything else. Oct 30, 2019 at 18:12
• @Bernard, could you help me out please? I'm stuck. :/^
– SFR
Oct 31, 2019 at 13:17