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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 $k \in \mathbb N.$

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  • $\begingroup$ I rolled back your deletion of your trials on (a) & (b) as they seemed to be demonstrate your effort in your questions. I suggest asking one question per post though. $\endgroup$ Dec 21 '20 at 9:51
  • $\begingroup$ @GNUSupporter8964民主女神地下教會 I already deleted that question as it was not accepted by the community. How did you return it? $\endgroup$
    – user838843
    Dec 22 '20 at 0:00
  • $\begingroup$ Those questions are always there. You had actually removed your own attempt on the question. I don't understand why you said they were not accepted by the community. One is always encouraged to share his/her own attempts on the question. The question (a) is referenced in the answer, so it would be odd to delete question (a). Questions are not a moving target. Once they have received an answer, any edits that invalidate existing answers should be discouraged. $\endgroup$ Dec 22 '20 at 8:58
  • $\begingroup$ @GNUSupporter8964民主女神地下教會 this is why I deleted the whole question after deleting my attempts .... I just wanted not be rejected by the community. $\endgroup$
    – user838843
    Dec 27 '20 at 4:53
  • $\begingroup$ Deleting your own question after receiving an answer is an abuse of the system. You have to wait for other users to vote on the answer. You may search for relevant guidelines are asking questions on this site on Math.Meta.SE. $\endgroup$ Dec 27 '20 at 21:03
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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.

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