(I have a couple of questions about graded ideals and I would appreciate any help/ideas anyone may have on the following. I had posted this one earlier, but nothing came of it, so deleted it, trimmed it a bit, and reposting it.)
Let $R = \mathbb{R}[x, y]$, and consider ideal $I = (a, b)$, with $a, b \in R$ homogeneous polynomials of degree $\bar{n}$.
Q1. If $\gcd(a, b) = 1$, then for sufficiently high degree $I_n = (x,y)^n$.
Q2. If $\gcd(a, b) = x^k$, then for sufficiently high degree $I_n = x^k(x,y)^{n-k}$.
My approach:
A1. I suppose that showing $x^iy^{n-i} \in I$, $0 \leq i \leq n$, for $n \gg 1$ will work, but not sure how to follow through with this idea.
A2. (Assuming that Q1 has been answered.)
$$ \gcd(a,b) = x^k \Rightarrow \gcd(a/x^k, b/x^k) = 1 $$ Then, for $m \gg 1$, $x^iy^{m-i} \in (a/x^k, b/x^k)$, $0 \leq i \leq m$. Thus, for $n \gg 1$, noting that elements of $I_n$ must be multiples of $x^k$, we have using Q1 that $I_n = x^k(x,y)^{n-k}$.
Any help would be appreciated, thank you.
Assumption:
$a, b$ are obtained by homogenization of univariate polynomials $a', b'\in \mathbb{R}[x] = R'$ with $\gcd(a', b') = 1$.
Solution
Then, for all $i \geq 0$, there exist $f_i', g_i' \in R'$ such that, $$ f_i'(x) a'(x) + g_i'(x) b'(x) = x^i. $$ Therefore, upon homogenization in degree $m \geq i$, $$ y^{m} f_i'(x/y) a'(x/y) + y^m g_i'(x/y) b'(x/y) = y^{m-i}x^i, $$ Then, since the degree of $a$ and $b$ is $\bar{n}$, and assuming that $p = \max_i(\max(\deg(f_i'), \deg(g_i')))$, we have for all $i \geq 0$, $$ y^{p + \bar{n}} \left(f_i'(x/y) a'(x/y) + g_i'(x/y) b'(x/y)\right) = y^{p+\bar{n}-i}x^i,\\ f_i(x,y)a(x,y) + g_i(x,y)b(x,y)=y^{p+\bar{n}-i}x^i, $$ where $f_i, g_i$ are the homogenizations of $f_i', g_i'$ in degree $p$. And therefore, for all $n \geq p+\bar{n}$, $I_n = (x, y)^n \cap R_n = R_n$.