Generators of an ideal in a polynomial ring This question has me stumped and I would appreciate any help you can offer!
Suppose that $I$ is an ideal of a polynomial ring $R[X]$, that $g$ is a monic element of $I$ and that ${\rm deg}(g) = n$. Put $P_n = \{f ∈ R[X]| {\rm deg} (f) \leq n − 1\}$. Prove that $I$ is generated by $g$ and the set $I\cap P_n$.
I feel a good way to start would be like this. Take $R[X]/(g)$. Note that this is isomorphic to $P_n$. We know $I/(g)$ is an ideal of $R[X]/(g)$. Where do I go from here?
 A: Welcome to MSE!
Pick your favorite $h \in I$. We want to show that we can write $h$ using $g$ and the $f$s in $P_n$.
Well, let's look at the degreee of $h$. If $\text{deg}(h) < n$, then we're golden! In this case $h \in P_n$ and we're already done.
What if $\text{deg}(h) \geq n$, though? We know we have a monic polynomial $g$. So if
$$h = h_N x^N + h_{N-1} x^{N-1} + \ldots + h_n x^n + h_{n-1} x^{n-1} + \ldots + h_0$$
we start out subtracting $h_N x^{N-n}g$. This kills off the $h_N x^N$ term of $h$, and gives us a new polynomial of degree $N-1$. We keep subtracting off the highest degre term until we hit $h_n x^n$. Now when we subtract $h_n x^{n-n} g = h_n g$, we're left with a polynomial of degree $n-1$, which is perfect! Because this means we're left with a polynomial in $P_n$.
So we've shown that
$$
h - h_N x^{N-n} g - h_{N-1} x^{N-1-n} g - \ldots - h_n g = f \in I \cap P_n
$$
and moving all the $g$ terms to the other side shows that $h \in \langle g, I \cap P_n \rangle$.
To make sure you understand the procedure (and to convince yourself of its correctness)

*

*Can you show explicitly that $5x^4 - 3x^3 + 2x^2 + 11x - 9$ is in the ideal generated by $x^2 + 1$ and $P_2$? Here I'm imagining $\mathbb{Z}[x]$
as our ring, so you aren't allowed to divide by anything!


*Now pick a weirder ring. Maybe $R = \mathbb{Z}[\epsilon]/(\epsilon^2 = 0)$ or something similar. If there's a weird ring that you like, you should use that one. Pick a fun (monic!) polynomial of degree $2$ or $3$, then pick a polynomial of degree $5$ or $6$ and again check that this works.


*What goes wrong if $g$ is not monic?

I hope this helps ^_^
