# Name for the reals augmented with an $x$ such that $x^2 = x$

If you add an $x$ such that $x^2=-1$ to the reals, you get the complex numbers. If you add an $x$ such that $x^2=0$ to the reals, you get the dual numbers. If you add an $x$ such that $x^2=+1$ to the reals, you get the split complex numbers.

If you add an $x$ such that $x^2=x$ to the reals, you get... what? What's the name for that number system? You know, the one where $e^{a+bx} = e^a + (e^{a+b} - e^a)x$.

• I believe this will be isomorphic to $\Bbb{R}\times \Bbb{R}$. Which incidentally is the same as the split complex numbers. – jgon May 22 '15 at 5:06
• I don't understand (sorry if it's just me), but don't $0, 1$ already satisfy that? – MCT May 22 '15 at 5:08
• @Soke, I was assuming we were taking $\Bbb{R}[x]/(x^2-x)$. – jgon May 22 '15 at 5:08
• @jgon Oh okay, the material is probably just going over my head. – MCT May 22 '15 at 5:10
• @Soke 0 and 1 do satisfy $v^2 = v$, but the question is about what you get when you add a third solution. – Craig Gidney May 22 '15 at 5:32

If $x^2=x$, then

$$(2x-1)^2=4x^2-4x+1=4x-4x+1=1$$

So you actually have the split-complex numbers in disguise.

• Doesn't this just show that we can embed the split complex numbers into the number system of interest? It doesn't show they're isomorphic. – goblin May 22 '15 at 5:11
• @goblin: They're both $2$-dimensional... – Micah May 22 '15 at 5:11
• It's pretty easy to find an inverse to show the embedding is onto and $1-1$ – Thomas Andrews May 22 '15 at 5:14
• Interesting, so it's just a basis change where $j$ = $2x-1$? Good enough for me! Thanks. – Craig Gidney May 22 '15 at 5:14
• indeed, adding $x$ so that $x^2+bx+c=0$, the resulting ring must be isomorphic to one of the three listed above, depending on whether $b^2-4c$ is negative, zero, or positive. – Thomas Andrews May 22 '15 at 5:19

Similarly to Micah's answer, but showing an ismorphism to $\Bbb{R}\times \Bbb{R}$, note that

Therefore the mapping to $\Bbb{R}\times \Bbb{R}$: $\varphi(a(1-x)+bx)=(a,b)$ preserves multiplication. Showing that it preserves addition is easy. Similarly the kernel must be trivial since if $\varphi(a(1-x)+bx)=0$, we have $a=b=0$, so $a(1-x)+bx=0$. Finally, this homomorphism is trivially surjective. Note that the mapping is well defined since $1-x$ and $x$ form a basis.