pdf with other exercises: [[20241121_DiskMat_Serie9.pdf]]
references

• https://crypto.stackexchange.com/a/41105 (vague, but simply explained)
• https://crypto.stackexchange.com/a/12472 (more details)
• https://www.aleksandrhovhannisyan.com/blog/modular-arithmetic-and-diffie-hellman/ (not overly helpful)

9.5)b)
Let $⟨R,+,-,0,\cdot ,1⟩$ be a ring and $a,b\in R$ arbitrary.
For $1-ab$ to be a unit, the following must be true for a $x\in R$:

Assume $x:=(1-ab{)}^{-1}$. Then condition $(1)$ is fulfilled, and $1-ab$ is a unit.
From $(1)$, we get:

We can also transform the second part of $(1)$:

From Lemma 5.17 (ii), we know that $(-a)b=-(ab)$ for some $a,b\in R$. We can use the lemma to show that $a(-b)=-(ab)$:

This means that

To show that $1-ba$ is a unit, we have to show that, for $y\in R$

According to the hint, we assume that $y:=1+bxa$. We have to show that

To show $(1-ba)(1+bxa)=1$:

To show that $(1+bxa)(1-ba)$:

Since $1-ab$ is a unit and $(1-ba)(1+bxa)=1$ and $(1+bxa)(1-ba)=1$, according to definition 5.20 (definition unit) $a-ba$ is also a unit $◻$ .