I don't really get this question....isn't $J$ actually $k[x_1,...,x_{n-1}]$ itself?
Let $I \subset k[x_1...x_{n-1},y]$ be an ideal. For a polynomial $f \in k[x_1,...x_{n-1},y]$, write $f=g_0+g_1y+...+g_my^m$ and denote $g_m \in k[x_1,...,x_{n-1}]$ the leading coefficient of $f$ with respect to $y$. Then show that $$J=\{g \in k[x_1...x_{n-1}]\ |\ \text{there is } f \in I \text{ such that } g \text{ is the leading coefficient of }f\},$$ is an ideal.
I am trying to prove it satisfies the definition of an ideal. I am stuck or confused where I have to show that, say $g_i,g_j \in J$ then $g_i-g_j \in I$. So it's closed under subtraction.
But I don't know, If $g_m \in k[x_1...x_{n-1}]$ then, isn't ANY possible polynomial in $k[x_1...x_{n-1}]$ I can dream of possible for any $g_i,g_j$? So well, their subtraction will clearly still be in $k[x_1...x_{n-1}]$ and....isn't $J = k[x_1...x_{n-1}]$ then?
I don't see any restriction on $g_m$ here, it can be every single possible polynomial in $k[x_1...x_{n-1}]$ yes? I don't see why not.
Then the proof seems...rather trivial. As long as $g_i-g_j \in k[x_1...x_{n-1}]$ which it clearly is, it is also in $J$.
Am I right?? I'm quite confused...because I feel like I'm wrong somewhere but i don't see where
Consider the ideal $I = (x y + 1)$ of the ring $k[x,y]$. Then the ideal $J$ consists only of $g(x) \in k[x]$ which are divisible by $x$. This follows, as every $h \in I$ is of the form
$$h = (x y + 1) (g_m(x)y^m + \cdots + g_1(x) y + g_0(x))$$
In general $J$ is an ideal: if $f_1 = g_1 y^{m_1} + r_1 \in I$ where every power of $y$ in $r$ is less than $m_1$ and similar $f_2 = g_2 y^{m_2} + r_2 \in I$ and without restricting generality $m_2 \geqslant m_1$, then $g_1 - g_2$ is the leading coefficient of $f_1 y^{m_2-m_1} - f_2$ considered in $k[x_1,\ldots,x_{n-1}][y]$.