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We consider the polynomial ring $R=\mathbb{R}[t]$ and the $R$-module $M=\mathbb{R}^3$, where $a\cdot x$ ($a\in R,x\in M$) is defined as usual if $a\in \mathbb{R}$, and $a\cdot x=(x_1, 0, 0)$ if $a=t, x=(x_1, x_2, x_3)$.

From the structure theorem for finitely generated $R$-module there is a list $p_1, p_2, \dots , p_n$ irreducible of $R$ and a list $k_1, k_2, \dots , k_n$ positive integers such that $$M\cong R/\langle p_1^{k_1}\rangle \oplus \dots \oplus R/\langle p_n^{k_n}\rangle$$

How can we find these lists?

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Hint: The polynomials $p_1,\dots,p_n$ must each annihilate a nontrivial submodule of $M$, so to identify these polynomials, try seeing if you can find elements of $\mathbb R[t]$ that annihilate some non-zero element of $M$.

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  • $\begingroup$ Which form have the elements of $M$ ? $\endgroup$
    – Mary Star
    Commented May 28, 2016 at 14:34
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    $\begingroup$ The elements of $M$ are just triples of real numbers. So, for example if we tried checking the action of the irreducible polynomial $p=2+t$ on $m=(1,2,3)$, we would get $p \cdot m = (2+t) \cdot (1,2,3) = 2 \cdot (1,2,3) + t\cdot(1,2,3) = (2,4,6)+(1,0,0) = (3,4,6)$. If the result had come out to $(0,0,0)$, then we would have identified $p$ one of the $p_i$ and $m$ as an element of one of the summands in the decomposition of the structure theorem. So try different values for $p$ and $m$ and see if you can find one that works. $\endgroup$ Commented May 28, 2016 at 14:39
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    $\begingroup$ It looks like you dropped the coefficient $a_i$ of $t^i$ for $i=1,\dots,n$. $\endgroup$ Commented May 28, 2016 at 15:25
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    $\begingroup$ Yes, that looks good. $\endgroup$ Commented May 28, 2016 at 19:17
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    $\begingroup$ More or less right. Now can you find specific choices of the $m_i$ and $p_i$ that work and are non-trivial, i.e., as in your cases (2) and (3)? $\endgroup$ Commented May 28, 2016 at 19:57

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