Let $F$ be a field and $R = F[X]$. Suppose $I$ is an ideal of $R$. Show that $I = (p(X))$ for some $p(X)$ in $F[X]$. (Hint: consider a polynomial $p(X)$ of least degree in $I$.)
I'm trying to do this question but can't seem to start. if anyone could give me a hint that would be appreciated.
My instinct was to take a monic polynomial (which exists since we are in a field) of degree $1$ and divide given the first two parts of the question were about polynomial division.
Thanks.
I think you need to be a bit more general then starting with a degree one polynomial. We are in a PID, so for example, in $\mathbb{Q}[x]$, $x^2 + 1$ generates a principle ideal, however it is not of degree one. This is my first impression, mind you it's been a long time since I studied this stuff, and this is just a first attempt, but should get you started.
Let $p\in I$ be as the hint suggests. Now suppose that $f$ is some polynomial in $I$ such that $p\not|f$, then $f = gp + r$ for some $r$ with degree of $r<p$. Contradiction as this tells us that $r=f-gp\in I$, but we picked $p$ to be of least degree.
I think the point here is that we are in a PID or a ED actually, so ideals are prime/generated by one element which is irreducible.