Jtdylktuy

Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$

Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...

Thanks!

Here's a beginning: “Suppose that $a,b in R^times$.”

Here's the end: “Therefore, $ab in R^times$.

The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?

With

$a, b in R^times, tag 1$

we have

$c, d in R^times tag 2$

with

$ac = bd = 1_R, tag 3$

where $1_R$ is the multiplicative identity of $R$; then

$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$

that is,

$ab, cd in R^times. tag 5$

$OEDelta$.

Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$

so $abin R^times $.