Show that ideals generated by leading coefficients of a polynomial ring are proper subsets of one another. 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 $d_1\leq d_2 \leq \dots \leq d_{n +1}$?
 A: Suppose that $i<j$, then $(f_1,\ldots,f_{i-1})\subset(f_1,\ldots,f_{j-1})$, and so
$$ f_j \in I\setminus(f_1,\ldots,f_{j-1})\subset I\setminus(f_1,\ldots,f_{i-1}). $$
By the minimality of the degree of $f_i$ in $I\setminus(f_1,\ldots,f_{i-1})$, we must have that $d_i\leq d_j$. Hence $d_1\leq d_2\leq \cdots d_{n+1}$.
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 $\deg(g)<d_{k+1}$.
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.
