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If $L$ is the submodule of $\mathbb{Q}[x]^{(3)}$ generated by $(2x-1,x,x),(x,x,x),(x+1,2x,x)$. How do we write $\mathbb{Q}[x]^{(3)}/L$ as a direct sum of cyclic modules?

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I'm tempted to say "the usual way". Same way one deals with finitely generated modules over any PID. – Arturo Magidin Dec 13 '11 at 3:43
do you mind elaborating? – tristan Dec 13 '11 at 3:47
Do you mind saying why you want to know? We like to know the sources of questions. We also like to know what you've tried on a problem or what your thoughts are, so that the answer does not re-invent the wheel. – Arturo Magidin Dec 13 '11 at 3:51
i really have no idea how should i tackle this question? should i put the generators into a matrix and reduce it some how to Smith normal form? and since Q is not finitely generated, can we write it as a direct sum of cyclic modules? – tristan Dec 13 '11 at 4:00
Huh? $\mathbb{Q}[x]$ is finitely generated as a $\mathbb{Q}[x]$ module: it's generated by a single element. $\mathbb{Q}[x]^3$ is finitely generated as a $\mathbb{Q}[x]$-module: it is generated by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. And, yes, Smith Normal Form will do it, though you can easily simplify it a bit ahead of time. – Arturo Magidin Dec 13 '11 at 4:02
up vote 3 down vote accepted

Yes, you can simply take the matrix that has the vectors as rows and use Smith Normal Form; if the Smith normal form is $$\left(\begin{array}{ccc} \alpha & 0 & 0\\ 0 & \beta & 0\\ 0 & 0 & \gamma \end{array}\right)$$ where $\alpha,\beta,\gamma\in\mathbb{Q}[x]$, then $\mathbb{Q}[x]^3/L$ will be isomorphic to $(\mathbb{Q}[x]/(\alpha)) \oplus (\mathbb{Q}[x]/(\beta)) \oplus (\mathbb{Q}[x]/(\gamma))$, which is a sum of cyclic $\mathbb{Q}[x]$-modules.

Here, you can actually save a bit of work by noting that $$ \left(\begin{array}{ccc} 2x-1 & x & x \\ x & x & x\\ x+2 & 2x & x \end{array}\right) \to \left(\begin{array}{ccc} x & x & x\\ x+2 & 2x & x\\ 2x-1 & x & x \end{array}\right) \to \left(\begin{array}{ccc} x& x & x\\ 2 & x & 0\\ x-1 & 0 & 0 \end{array}\right)$$ so $L$ is generated by $(x-1,0,0)$, $(2,x,0)$, and $(x,x,x)$.

We can consider $\mathbb{Q}[x]^3$ as having basis by $(1,1,1)$, $(1,0,0)$ and $(0,1,0)$. If we do that, then the third generator of $L$ is just $x$ times the first generator of $\mathbb{Q}[x]^3$, and the submodule generated by the other two are in the direct summand generated by the second and third elements of the basis. So $$\frac{\mathbb{Q}[x]^3}{L} \cong \frac{\langle (1,1,1)\rangle}{\langle x(1,1,1)\rangle} \oplus \frac{\langle (1,0,0),(0,1,0)\rangle}{\langle (x-1,0,0), (2,x,0)\rangle}.$$ The first summand is isomorphic to $\mathbb{Q}[x]/(x) \cong \mathbb{Q}$ via the map that sends $x$ to $0$ (and $\mathbb{Q}$ is a cyclic $\mathbb{Q}[x]$-module, where $x$ acts as multiplication by $0$).

This reduces the problem to just trying to write $\mathbb{Q}[x]^2/M$ as a direct sum of cyclic modules, where $M$ is the submodule generated by $(x-1,0)$ and $(2,x)$. If nothing obvious suggests itself at this point, you can use the Smith Normal Form to see what this quotient looks like.

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thank you for your answer, it helps a lot. – tristan Dec 13 '11 at 4:19

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