Proof $x\in R^\times \wedge b\in R^\times \Rightarrow ab\in R^\times$

Question

Let $R$ be a communitative ring. Prove $a\in R^\times \wedge b\in R^\times >\Rightarrow ab\in R^\times$ with $R^\times := \{x\in R\ |\ x\ \text{ >invertible}\}.$

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!

Answer 1

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?

Answer 2

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,

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

$OE\Delta$.

Answer 3

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

so $ab\in R^\times $.