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I have to calculate the invariant factor decomposition of $M$ the $\mathbb Q[x]-$module with generator $e_1$ and $e_2$ and the relation $$x^2e_1+(x+1)e_2=0$$ $$(x^3+2x+1)e_1+(x^2-1)e_2=0.$$ So, I need to calculate the Smith normal form of the matrix $$\begin{pmatrix}x^2&x^3+2x+1\\ x+1& x^2-1\end{pmatrix}.$$

But in the calculation, we have that $$\begin{pmatrix}x^2&x^3+2x+1\\ x+1& x^2-1\end{pmatrix}\underset{(*)}{\to} \begin{pmatrix}-x&3x+1\\ x+1&x^2-1\end{pmatrix} \to ...\to \begin{pmatrix}1&0\\ 0& (x+1)^3\end{pmatrix},$$ but I don't understand the step $(*)$. Why $$(x^2,x+1,x^3+2x+1)=(-x)\ \ ?.$$

Other question By the way, in my correction it's written that $M\cong \mathbb Q[x]/(x+1)^3$. Shouldn't it be $M\cong \mathbb Q[x]\oplus \mathbb Q[x]/(x+1)^3$ ?

Algorithm added

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1 Answers 1

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Simply by this elementary row operation: $\mathrm R_1\leftarrow \mathrm R_1 -x\,\mathrm R_2$.

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    But in my algorithm, it's written that : first we have to be sure that at the place (1,1), we have the generator of $(x^2,x+1,x^3+2x+1)$. So, why $-x$ generate this ideal ? – 2017-01-27
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    I don't this algorithm, I know another, based on Bézout's identity. Do you have a link? – 2017-01-27
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    Yes, page 12 and 13 [Here](http://moodle.epfl.ch/pluginfile.php/1756811/mod_resource/content/2/notes.pdf) – 2017-01-27
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    Unfortunately, I'm not allowed to connect. Could post a link to the relevant extracted pages from this pdf? – 2017-01-27
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    I added to my original post. Thank you for helping :) – 2017-01-27
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    That's what I know. I has nothing to do with a generator of the ideal (x^2, x+1, x^3+2x+1)$ – 2017-01-27
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    Ok, thank you. And for my other question (last question in my post), do you agree with me ? – 2017-01-27
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    No: $M\simeq \mathbf Q[x]/(1)\times\mathbf Q[x]/(x+1)^3 \simeq\{0\}\times\mathbf Q[x]/(x+1)^3\simeq\mathbf Q[x]/(x+1)^3$. – 2017-01-27
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    If so, in a previous exercise, I have M generated by $(e_1,...,e_4)$ s.t. $e_1-2e_3+2e_3-e_4=0$, $9e_1-6e_2-8e_3+e_4=0$ and $e_1+2e_3+5e_4=0$. Then I get the matrix $\begin{pmatrix}1&9&1\\-2&-6&0\\2&-8&2\\ -1&2&5\end{pmatrix}$, and thus the smith normal form $$\begin{pmatrix}1&0&0\\0&2&0\\0&0&26\\ 0&0&0\end{pmatrix}$$ and the say that $M\cong \mathbb Z\oplus \mathbb Z/2\mathbb Z\oplus \mathbb Z/26\mathbb Z$. Why here is it different ? By what you said it should be $\mathbb Z/2\mathbb Z\oplus \mathbb Z/26\mathbb Z$, no ? – 2017-01-27
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    If you have relations between the $e_i$s, this means you work in the *dual* of $\mathbf Z^4$, and the matrices should be transposed. So the Smith normal form should be written as $\begin{pmatrix}1&0&0&0\\0&2&0&0\\0&0&26&0\end{pmatrix}$, $M$ is isomorphic to $\mathbf Z/(1)\times\mathbf Z/(2)\times\mathbf Z/(26)\times\mathbf Z/(0)\simeq \{0\}\times\mathbf Z/(2)\times\mathbf Z/(26)\times\mathbf Z\simeq \mathbf Z/(2)\times\mathbf Z/(26)\times\mathbf Z$. – 2017-01-27