Show that ideals generated by leading coefficients of a polynomial ring are proper subsets of one another.

Question

Suppose I is an ideal of $R[x]$. Let $f_1$ be a member of $I\setminus (0)$ such that $n_1=\deg(f_1)$ is minimal, inductively let $f_{n+1}$ be a member of $I\setminus (f_1,\dots,f_n)$, ie in $I$ but not in the ideal generated by $f_1$,$f_2$,$\dots$,$f_n$ such that $d_{n+1}=\deg(f_{n+1})$ is minimal. Let $a_i$ be the leading coefficient of each $f_i$.

How can I show that $(a_1)$, the ideal generated by $a_1$, is strictly contained in $(a_1, a_2)$ is strictly contained in $(a_1, a_2, a_3 )$ ... is strictly contained in $(a_1, a_2, a_3, \dots, a_{n+1} )$? How can I also also show $d_1\leq d_2 \leq \dots \leq d_{n +1}$?

Answer 1

Suppose that $i

Suppose that for some $k$, the inclusion $(a_1,\ldots, a_k)$ in $(a_1,\ldots,a_{k+1})$ is not strict. Then $a_{k+1}$ is in $(a_1,\ldots,a_k)$, and so

$$a_{k+1} = r_1a_1 + \cdots + r_ka_k$$

for some $r_1,\ldots r_k\in R$. Define $g\in I$ by

$$g(x)=f_{k+1}(x) - r_1x^{d_{k+1}-d_1}f_1(x) - \cdots - r_kx^{d_{k+1}-d_k}f_k(x).$$

Notice that the degree of all the summands are $d_{k+1}$, hence $\deg(g)\leq d_{k+1}$. However, the coefficient of $x^{d_k}$ in $g(x)$ is $a_{k+1} - r_1a_1 - \cdots - r_ka_k=0$, so in fact $\deg(g)

Moreover $g\notin(f_1,\ldots,f_k)$, as otherwise

$$f_{k+1}(x) = g(x) + r_1x^{d_{k+1}-d_1}f_1(x) + \cdots + r_kx^{d_{k+1}-d_k}f_k(x)$$

would be in $(f_1,\ldots,f_k)$ also, a contradiction. So $g\in I\setminus(f_1,\ldots,f_k)$, and $g$ has smaller degree than $f_{k+1}$. This contradicts our choice of $f_{k+1}$, hence the assumption that $(a_1,\ldots,a_k)$ was not strictly contained in $(a_1,\ldots,a_{k+1})$ must be false.