# Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.

Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.

Here is what I have:

$a+b =b+a$ since $R$ is commutative.

Now,

$$(b+a) \cdot b^{-1} = 1+ab^{-1} \\ a^{-1} \cdot (1+ab^{-1}) = a^{-1} + b^{-1}$$

Thus, $a^{-1} + b^{-1} =a^{-1} \cdot (a+b) \cdot b^{-1}$

Therefore, $(a^{-1} + b^{-1})^{-1} = b \cdot (a+b)^{-1} \cdot a$

And thus, $a^{-1} + b^{-1}$ is a unit in $R$ as well.

Does my answer sound logical. Or are there errors in it?

## 2 Answers

Your answer works perfectly fine. In fact, your answer works in non-commutative rings as well.

When we say that a ring is commutative, we mean $ab = ba$. We will have $a + b = b + a$ is any ring.

• How then is the fact that this ring is commutative relevant in this problem? – user198504 Sep 14 '15 at 14:41
• In fact, it's irrelevant. It does make the proof a bit easier, though, since we can now write $ab(a^{-1} + b^{-1}) = a + b$. – Omnomnomnom Sep 14 '15 at 14:45
• Well thanks for reminding me that commutativity in rings refer to commutative multiplication. My brain really floats away from me sometimes. I always knew that rings are abelian groups as well :/ – user198504 Sep 14 '15 at 14:48

Your proof is fine.

One way to discover it is by computing freely: $$\frac{1}{\dfrac{1}{a}+\dfrac{1}{b}} = \frac{1}{\dfrac{a+b}{ab}} = \frac{ab}{a+b}$$