Here is the question I want to answer letter $(c)$ of it:
Let $R$ be a commutative ring, and let $R[x] = R^{[1]}.$ Given an ideal $I \subset R[x]$ and $k \in \mathbb N,$ let $L_k(I) \subset R$ consists of $0$ and all leading coefficients of degree$-k$ polynomials of $I.$ Given ideals $I \subset J$ of $R[x],$ show the following.
$(a)$ $L_k(I)$ is an ideal of $R$ for each $k \in \mathbb N.$
$(b)$ $L_k(I) \subset L_k(J)$ for each $k\in \mathbb N.$
$(c)$ $L_k(I) \subset L_{k + 1}(I)$ for each $k\in \mathbb N.$
Here is my trial for letter $(c)$:
Let $x_{0} \in L_k(I)$ and we want to show that $x_{0} \in L_{k + 1}(I)$ for each $k\in \mathbb N.$
Now, since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of some degree$-k$ polynomial of $I,$ name it $p(x).$ Now, since $I$ is an ideal of $R[x],$ then $xp(x) \in I$ as $x \in R[x].$ If $p(x) = x_{0}x^k + a_{k-1}x^{k-1} + \dots + a_1x +a_0,$ then $xp(x) = x_0 x^{k+1} + a_{k-1}x^{k} + \dots + a_1x^2 +a_0x + 0,$ i.e. $xp(x)$ is a polynomial of degree $k+1$ with leading coefficient $x_0,$ which by definition of $L_{k + 1}(I)$ means that $x_{0} \in L_{k + 1}(I).$ Since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$
Is my trial correct? or should I use induction in the proof? could anyone clarify this to me please?
EDIT:Also, here are my trials for the solutions of $(a)$ and $(b)$ and I have the same question: Should I use induction in any of the following solutions:
$(a)$
Given an ideal $I \subset R[x]$ and $k \in \mathbb N,$ Define $L_k(I) \subset R$ as consisting of $0$ and all leading coefficients of degree$-k$ polynomials of $I.$ We want to prove that $L_k(I)$ is an ideal of $R$ for each $k \in \mathbb N.$
\textbf{(1) Showing that $L_k(I)$ is a subgroup of $(R, +)$ for each $k \in \mathbb N.$}
$0_R$ is in $L_k(I)$ by its definition.
Now, let $x_{0}, y_{0} \in L_k(I),$ we want to show that $x_{0} - y_{0} \in L_k(I).$
Since $x_{0}, y_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $p(x)$ and also, $ y_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $q(x).$ Now, since $I$ is an ideal, then the polynomial $p(x) - q(x) \in I$ which is a degree$-k$ polynomial of $I$ with leading coefficient $x_{0} - y_{0}.$ Now, by the definition of $L_k(I),$ we have that $x_{0} - y_{0} \in L_k(I)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$
So the previous 3 paragraphs implies that $L_k(I)$ is an abelian subgroup of $(R, +)$ as $R$ is commutative.
\textbf{(2) Showing that for $x_{0} \in L_k(I),$ and $r \in R,$ we have that $rx_{0} \in L_k(I),$ for each $k \in \mathbb N.$}
Let $x_{0} \in L_k(I),$ and $r \in R.$ Since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $p(x).$ Now, since $I$ is an ideal, then the polynomial $rp(x) \in I$ which is a degree$-k$ polynomial of $I$ with leading coefficient $rx_{0}.$ Now, by the definition of $L_k(I),$ we have that $rx_{0} \in L_k(I)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$
So, from $(1)$ and $(2),$ we get that $L_k(I)$ is an ideal of $R.$
$(b)$
Assume that $I,J$ are ideals of $R[x]$ such that $I \subset J.$
First, $0_{R}$ is in both $L_k(I)$ and $L_k(J)$ for each $k\in \mathbb N$ by their definitions.
Second, assume that $x_{0} \in L_k(I)$ and we want to show that $x_{0} \in L_k(J)$ for each $k\in \mathbb N.$
Now, since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of some degree$-k$ polynomial of $I,$ name it $p(x).$ But, by the given, we have that $I \subset J,$ then $p(x)$ is also a degree$-k$ polynomial in $J$ that has a leading coefficient $ x_{0}.$ So, by definition of $L_k(J)$ we have that $x_{0} \in L_k(J)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$
I can't imagine that induction is necessary here. We must simply fix a non-negative integer $k,$ and show that each of the claims holds for that $k.$ Considering that $k$ was chosen arbitrarily, each of the claims will hold for each non-negative integer $k,$ and our proof is complete.
I notice that there is a more succinct way of completing these proofs. Particularly, for any commutative ring $R,$ a nonempty subset $I$ of $R$ is an ideal of $R$ if and only if for any elements $i, j \in I,$ we have that (1.) $i - j \in I$ and (2.) $ri \in I$ for any element $r \in R.$ By the one-step subgroup test, the first shows that $(I, +)$ is a subgroup of $(R, +);$ the second shows that $I$ is closed under multiplication by elements of $R.$ Combined, these say that $I$ is an ideal of $R.$
Bearing this in mind, the proof of point (a.) should go as follows.
Proof. Given any non-negative integer $k,$ observe that $L_k(I)$ is nonempty by hypothesis that $0_R$ is in $L_k(I).$ Given any two elements $r, s \in L_k(I),$ therefore, there exist polynomial $p_r(x)$ and $p_s(x)$ of degree $k$ in $I$ whose leading coefficients are $r$ and $s,$ respectively. Consequently, the leading coefficient of the degree $k$ polynomial $p_r(x) - p_s(x)$ is $r - s,$ hence we have that $r - s \in L_k(I).$ Further, for any element $t$ of $R,$ we have that $tr = 0_R$ or $tr \neq 0_R.$ For the case of the former, it follows that $tr = 0_R$ is an element of $L_k(I),$ and for the latter, we have that $tp_r(x)$ is a polynomial of degree $k$ with leading coefficient $tr.$ Either way, it follows that $L_k(I)$ is closed under multiplication by elements of $R.$ We conclude that $L_k(I)$ is an ideal of $R.$ Considering that $k$ is arbitrary, $L_k(I)$ is an ideal of $R$ for all non-negative integers $k.$ QED.