Show that $I(a,b)=I(a',b')$

Question

Please help me to solve this problem:

"Let $a,b,a',b',m,n,r,s$ be integers such that $m.s-n.r=1$ or $m.s-n.r=-1$, $a'=m.a+n.b$ and $b´=r.a+s.b$. Show that $I(a,b)=I(a',b')$, where $I$ is the symbol for an ideal, i.e., $I(a,b)=$ $ \left\{ na+mb;m,n \in A \right \}$, $A$ is a ring and $I \subset A$".

I know that $I(a,b)=I(b,a)$ and $I(a,b)=I(a,b-ta), t\in A$.

I also know that

$ra'=rma+rnb$

$mb'=mra+msb$

Which implies $b=mb'-ra'$.

Similarly, $nb'=nra+nsb$ and $sa'=sma+snb$ implies $a=sa'-nb'$.

Then, $I(a,b)=I(sa'-nb',mb'-ra')$, but I can't go any further.

Answer 1

Hint $\ \ \left[\matrix{a'\\ b'}\right] = \left[\matrix{m & n\\ r & s}\right] \left[\matrix{a\\ b}\right] \,\Rightarrow\, a',b'\,\in I(a,b)\,\Rightarrow\, I(a',b')\subseteq I(a,b)$

The determinant $=\pm1\,$ is invertible so, by Cramer, we can solve for $\,a,b\,$ as an $R$-linear combination of $\,a',b',\,$ so a similar argument yields the reverse containment $\,I(a,b)\subseteq I(a',b')$

This clearly generalizes to show that ideals are preserved by such linear transformations with imvertible (unit) determinant, i.e. by unimodular transformations of their generators (basis).

Generally note $\,\ (a_1,\ldots,a_j) \subseteq (b_1,\ldots,b_k)\iff a_1,\ldots,a_j \in (b_1,\ldots,b_k)\iff $

$$\iff \begin{bmatrix} a_1\\ \vdots\\ a_j\end{bmatrix}\,=\,\begin{bmatrix}c_{11} &\ldots &c_{1k}\\ \vdots &\ddots & \vdots\\ c_{j1}&\ldots &c_{jk} \end{bmatrix}\begin{bmatrix} b_1\\ \vdots\\ b_k\end{bmatrix}\ \ \text{for some }\ c_{i,j}\in R$$