Show that two homogeneous polynomials in two variables which share no linear factors generate power of maximal ideal I'm trying to show that given $g$ and $h$, homogeneous polynomials in $k[x,y]$ of degrees $m$ and $n$, respectively, if $g$ and $h$ share no common linear factors, then they generate the maximal ideal of $k[x,y]$ to the $m+n-1$'st power. It is clear that they generate some power of the maximal ideal from the nullstellensatz, and it is clear that if $g = x^m$ and $h = y^n$ that they generate all monomials of degree $m+n-1$, which form a basis for $\mathfrak{m}^{m+n-1}$. However, in general I'm having trouble showing that this is true (but I suspect that it is).
$\mathbf{Edit}$: I'm trying to do exercise 5.14b in Hartshorne, and I believe this is what I have to show to prove the statement in the exercise; if perhaps I've gone astray in coming to this conclusion, please point me in the right direction.
 A: I'm not an algebraist but I guess this result should be well known. Nevertheless, I found an elementary proof.
I don't assume $k$ algebraically closed. Set $I$ the ideal of $k[x, y]$ generated by
$g$ and $h$. We want to prove that $I \cap \mathfrak{m}^{m+n-1} = \mathfrak{m}^{m+n-1}$.
First, we will prove $x^{m+n-1} \in I$ and $y^{m+n-1} \in I$.
Let $G$ and $H$ the two polynomials of $k[z]$ such that
$$y^m G\Big(\frac{x}{y}\Big) = g(x,y)\ \mbox{ and }\ y^n H\Big(\frac{x}{y}\Big) = h(x, y).$$
Then $\mathrm{deg}\, G = \mathrm{deg}_h\, g$ and $\mathrm{deg}\, H = \mathrm{deg}_h\, h$ and $G$ and $H$ are relatively prime since $g$ and $h$ have no common linear factors. Thus according to the Bézout's identity, there exist $U, V \in k[z]$ such that
$$ U G + V H = 1 \mbox{ and } \mathrm{deg}\, U < \mathrm{deg}\, H,\ \ \mathrm{deg}\, V < \mathrm{deg}, G.$$
But $u = y^{n-1} U(\frac{x}{y}) \in k[x, y]$ and $v = y^{m-1} V(\frac{x}{y}) \in k[x, y]$. Hence
$$ y^{m+n-1} = u g + v h.$$
In other words, $y^{m+n-1} \in I$. By arguing with $y$ replaced by $x$, we got $x^{m+n-1} \in I$.
Now we proceed by induction on $m+n$. Assume the statement holds for any homogeneous polynomials of degree $s \leqslant m$ and $t \leqslant n$  such that $s + t \leqslant m+n-2$. If $x$ divides $g$ then $g = x f$ with $f \in k[x, y]$ homogeneous of degree $m-1$. Thus by induction hypothesis,
$$\langle f, h \rangle \cap \mathfrak{m}^{m+n-2} = \mathfrak{m}^{m+n-2}.$$
Hence for any $p, q$ such that $p+q = m + n-2$, we can find $u, v \in k[x, y]$ such that
$$ x^p y^q = u f + v h.$$
So, we get
$$ x^{p+1} y^q = u x f + x v h = u g + x v h.$$
We proved that $x^p y^q \in I$ for any $p \geqslant 1$ and $q \geqslant 0$ such that $p + q = m + n - 1$. Since $y^{m+n-1} \in I$ by the above argument, we have done in this case. Arguing similarly, we deal with the case $y \mid g$, $x \mid h$ and $y \mid h$. The last case is when $x$ and $y$ don't divide $g$ and $h$. Thus, there exist $g_1, h_1 \in k[x, y]$ homogeneous such that
$$ g = x^m + y^m + xy g_1\ \mbox{ and }\ h = x^n + y^n + xy h_1.$$
We can assume that $m \geqslant n$. If $m = n$ set $f = g - h$, otherwise set $f = g - (x^{m-n} + y^{m-n})h$. Then
$$ f = xy g_1 - xy h_1\ \mbox{ or }\ f = xy g_1 - x^{m-n}y^n - y^{m-n}x^n - x^{m-n+1}y h_1 - y^{m-n+1}x h_1.$$
So $x$ and $y$ divide $f$, $f$ and $h$ don't share a linear factor and $I = \langle f, h \rangle$. We apply the above argument to conclude.
Remark 1: I don't know if it is useful but the proof is completely effective. You can write easily an algorithm.
Remark 2: It could be interesting to investigate the higher dimension case. The following statement might hold:

Let $f_1, \ldots, f_n \in k[x_1, \ldots, x_n]$ be homogeneous polynomials of degree $m_1, \ldots, m_n$ without common zero in $\bar{k}^n-\{0\}$. Then
$$ \langle f_1, \ldots, f_n \rangle \cap \mathfrak{m}^{m_1 + \cdots + m_n - n + 1} = \mathfrak{m}^{m_1 + \cdots + m_n - n + 1}.$$

Edit:
The above proposition is a consequence of properties of the Macaulay resultant. Indeed, if $f_1, \ldots, f_n \in k[x_1, \ldots, x_n]$ are homogeneous polynomials, then we can prove
$$ \mathrm{Res}(f_1, \ldots, f_n) x_1^{p_1} \cdots x_n^{p_n} \in \langle f_1, \ldots, f_n \rangle, $$
for any $p_1, \ldots, p_n \in \mathbb{N}$ such that $p_1 + \cdots + p_n = m_1 + \cdots + m_n - n + 1$ and $\mathrm{Res}(f_1, \ldots, f_n) \in k$ is the Macaulay resultant. But the Macaulay resultant vanishes if and only if the system has a non-zero solution.
