2
$\begingroup$

There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe).

Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the relations $4x+y+2z=0, 5x+2y+z=0,6y-6z=0$. Determine the invariant factors of $M$.

My attempt:
I wasn't sure what theorem or reasoning to use, but I have a feeling to write the equations in matrix form and find the equivalent matrix. My question is normally if we write in matrix form, we should be priorly given a map or homomorphism, but this question did not, or whether somehow we can interpret the equations as map? if so, what is the map?

Let me try to make one. Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^3$ and $N=\langle(4,5,0),(1,2,6),(2,1,-6)\rangle\le\mathbb{Z}^3$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^3$ given by $(x,y,z)\to x(4,5,0)+y(1,2,6)+z(2,1,-6)$. Am I correct? Please correct me if I am wrong.

Then I find the equivalent matrix:
$\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix}$ $4R_2-5R_1$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 6 & -6\end{bmatrix}$ $R_3-2R_2$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $3R_1+R_2$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $R_2+R_3$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$ $R_1-2R_2$ $\begin{bmatrix}12 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$. Then by performing column and row swaps, the matrix becomes $\begin{bmatrix}3 & 0 & 0\\0 & 6 & 0\\0 & 0 & 12\end{bmatrix}$. Since 3,6,12 are non-units in $\mathbb{Z}$, then the invariant factors of M are 3, 6, 12.

However, the solution says that the invariant factors are 1,3,6.

Was my method correct? Where did I make the mistake?

Thanks!

$\endgroup$
3
$\begingroup$

When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by $$ \begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}. $$ Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced the spurious factor $12$.

I don't have the time to post the correct calculations (which are a lengthier alternative to the use of the determinant in this case), perhaps later.


So here's how you do it: $$\begin{align} &\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix} C_1 \leftrightarrow C_2 \begin{bmatrix}1 & 4 & 2\\2 & 5 & 1\\6 & 0 & -6\end{bmatrix} R_2 - 2 R_1 \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\6 & 0 & -6\end{bmatrix} R_3 - 6 R_1\\& \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} C_2 - 4 C_1, C_3 - 2 C_1 \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} R_3 - 8 R_2\\& \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & 0 & 6\end{bmatrix} C_3 - C_2 \begin{bmatrix}1 & 0 & 0\\0 & -3 & 0\\0 & 0 & 6\end{bmatrix} {-R_2} \begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix} \end{align}$$

$\endgroup$
  • $\begingroup$ @user71346, I have added the calculations to my answer. $\endgroup$ – Andreas Caranti May 22 '13 at 10:59
  • $\begingroup$ Thanks Andreas. That's amazing. Having seen your calculation, especially your final step when you multiply R2 by -1, so can I do the same thing by multiplying R3 of my matrix by 1/12? and then do the final rearrangement. $\endgroup$ – user71346 May 23 '13 at 3:10
  • $\begingroup$ Also thanks for suggestion, I will have a read at Jacobson's Basic Algebra I. $\endgroup$ – user71346 May 23 '13 at 3:11
  • $\begingroup$ @user71346, you're welcome. You can only multiply rows and column by an invertible element (and you are working over the integers here). $\endgroup$ – Andreas Caranti May 23 '13 at 6:23
1
$\begingroup$

There is a bit of theory to go through to explain how the invariant factors of the matrix you wrote down is actually related to $M,$ you will have to go through your notes to see the precise connection.

As for the calculation, perhaps recheck your operations, or post them here so we might we able to say where you went wrong.

It turns out that for this matrix, we can get away with doing no work by using the fact that the product of the invariant factors is equal to the determinant of the matrix, which we can compute is $18 = 2\cdot 3^2.$ Since each invariant factor must divide the next, the only possibility is $1,3,6.$

$\endgroup$
  • $\begingroup$ Thanks for the answer. I have added my matrix calculations. I have double checked, I don't think the mistakes are in the calculations? $\endgroup$ – user71346 May 22 '13 at 3:57
  • $\begingroup$ Ragib Zaman. I had edited the question, I hope it is more clear now. Where did I go wrong? Also, is there any other way without using the determinant? Many thanks. $\endgroup$ – user71346 May 22 '13 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.