Inverse of The Element of Ring

Question

Let $R$ be a ring with unity $x \in R$ such that $x^2 = x$. The inverse of $2x - 1$ is $\cdots$

This problem is rather easy to be tricky. I found that $2x-1$ is an inverse of itself because $(2x-1)(2x-1) = 4x^2-4x+1$ and by using the equation $x^2 = x$, I get $(2x-1)(2x-1) = 1$. So here I get. Instead by using this approachment, are there any precise ways to solve the problem?

Answer 1

As you noticed, the fact that $x^2 = x$ is leading us to think that $(2x-1)^{-1}$ is a polynomial of $x$ ; and such polynomial is of the form $ax+b$.

But - assuming commutativity - $$\begin{array}{rr} &(2x-1)(ax+b) = 1 \\ \iff & 2ax + 2xb -ax -b=1 \\ \iff & x(a+2b) -b = 1 \\ \iff & a = 2 \text{ and } b=-1 \end{array}$$

EDIT : since I did not make this clear enough, note that the last "$\iff$" is only valid because of the uniqueness of $(2x-1)^{-1}$

Answer 2

Precise way? No.

Fun way? Yes!

\begin{align} (2x-1)^{-1} & = -1 \cdot\dfrac{1}{1-2x} \\ & = -1 \cdot(1 + 2x +(2x)^2 + (2x)^3 + \ldots) \\ & = -1 \cdot(1 + 2x +2^2x + 2^3x + \ldots) \\ & = -1 \cdot(-x + 1 + x + 2x +2^2x + 2^3x + \ldots) \\ & = -1 \cdot(-x + 1 + x (1 + 2 +2^2 + 2^3 + \ldots)) \end{align} Uh-oh, we've got an infinite sum of the powers of $2$. Let's figure out its value. Set $$ a= 1 + 2 +2^2 + 2^3 + \ldots $$ Then $$a - 2a = (1 + 2 +2^2 + 2^3 + \ldots) - (2 +2^2 + 2^3 + 2^4 +\ldots) = 1,$$ so $a = -1$. Plugging this back into our original equation \begin{align} (2x-1)^{-1} & = \ldots \\ & = -1 \cdot(-x + 1 + x (1 + 2 +2^2 + 2^3 + \ldots)) \\ & = -1 \cdot(-x + 1 + x\cdot -1) \\ & = -1 \cdot(1 -2x) \\ & = 2x - 1 \end{align}

All of this, of course, is highly suspect and completely invalid nonsense. However, when you're dealing with a situation where you need to come up with an answer and it's enough to verify that the answer satisfies some condition, (like you did by checking that $(2x-1)(2x-1) = 1$), that trick of expanding $(1-x)^{-1}$ as an infinite power series can often get you to the right answer.

If you look at some of Euler's work he often plunges ahead with infinite sums completely unconcerned about what they actually are or might mean. Euler was a genius and we aren't, so we have to be much more careful and need some way to verify things at the end, but sometimes we can borrow the seven league boots of a genius and get away with it.

Answer 3

Not an answer to your question (there already are valid answers), but I feel that some geometric intuition would be interesting.

In the algebra of linear operators, idempotents (elements satisfying $x^2=x$) are precisely projection operators. Now, what is $2x-1$? Rewrite as $2x-1=x-(1-x)$. If $x$ is a projection, then $1-x$ is the complementary projection: its image is the kernel of $x$, and its kernel is the image of $x$. Any vector $v$ can be decomposed as a sum of its projections: $$v = 1\cdot v = (x + (1-x)) \cdot v = x \cdot v + (1-x) \cdot v,$$

while the operator $2x-1$ acts as

$$x \cdot v - (1-x) \cdot v$$

So, $2x-1$ negates the $(1-x)$-component of a vector, which is just a reflection through the subspace $\operatorname{Im} x$. And, of course, reflections $r$ always satisfy $r^2=1$, since reflecting twice does nothing. In this case, $r=2x-1$.