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Let $K$ be a field, let $R=K[x]$, and let $A\ne 0 \in R$. If $m > \deg(A) $, it is to show that there are no ideals $I_{1}=AR, I_{2},\ldots,I_{m}$ of $R$, all different from each other, such that $I_{1} \subset I_{2} \subset \cdots\subset I_{m}$.

This exercise is from a algebra book and I want to do it, but I don't see how to begin.

Does anybody see the beginning?

Tell me . Please.

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    I've tried to make the title more informative, and improved the LaTeX a bit. As has been mentioned to you before, **do not sign your posts**. It is against the policy of the site, and it will be removed.2011-11-01

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You can do this by hand in the polynomial ring but just let me point out a different way to do it that doesn't require knowing $K[x]$ is a PID.

Look at R' := K[x]/(A(x)). Then, the ideals of R' are in bijection with the ideals $I$ of $R$ such that $I \subset (A(x))$, the bijection being given by taking an ideal $I \subset (A(x))$ and looking at $I + (A(x))$ in the quotient. Notice that R' is a vector space over $K$ of dimension $\deg A$. Thus, any chain of ideals contained in $(A(x))$ gives a chain of subspaces of of R'. Such a chain cannot be longer than the dimension = $\deg A$.

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    I like that! Good one!2011-11-01
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I assume you want the ideals to be proper. For $(x)\subsetneq \mathbb{R}[x]$. Ok, so you know that since $k$ is a field that $k[x]$ is a PID. So, what you're saying is that $(A)\subsetneq (B_1)\subsetneq\cdots\subsetneq (B_{m-1})$. This says to me that $B_1$ properly divides $A$, and so $A=B_1 q_1(x)$ for some $q_1(x)\notin k$. Similarly $B_1=B_2 q_2(x)$ so that $A=B_2 q_1(x) q_2(x)$. Continuing in this way you find that $A=B_{m-1}q_1(x)\cdots q_{m-1}(x)$ and so this clearly implies by looking at degrees that $B_{m-1}\in k$ and so $I_{m-1}=k[x]$ contradictory to assumption.