Another ring associated with an idempotent (besides the corner ring)

Question

Given any ring $R$ (not necessarily with unity) and any idempotent $e$ in $R$, the corner ring $eRe$ is a ring with $e$ as the multiplicative identity.

Now consider the set $S = \{x \in R \mid ex = xe = e\}$. Define the addition, zero, additive inverse, and multiplication in $S$ as follows:

• The "sum" of $x$ and $y$ in $S$ is $x+y-e$. (This is in $S$ because $e(x+y-e) = ex+ey-ee = e+e-e = e$, and likewise, $(x+y-e)e = xe+ye-ee = e+e-e = e$.)
• The "zero" in $S$ is $e$. (This is in $S$ because $e$ is idempotent.)
• The "additive inverse" of $x$ in $S$ is $2e-x$. (This is in $S$ because $e(2e-x) = 2ee-ex = 2e-e = e$, and likewise, $(2e-x)e = 2ee-xe = 2e-e = e$.)
• The product of $x$ and $y$ in $S$ is the same as their product in $R$. (This is in $S$ because $exy = ey = e$, and likewise, $xye = xe = e$.)

Two questions:

Do the above operations make $S$ into a ring (with the same unity as $R$ if $R$ has one)? Also, is $S$ isomorphic to the corner ring $(1-e)R(1-e)$ if $R$ is a ring with unity?

Answer 1

Yeah, it’s a ring and it’s isomorphic to the corner ring, assuming that $R$ has a unity $1$. You can spare all tedious calculations by interpreting $S$ as the pullback of an additive subgroup $T ⊆ R$ by a translation by $e$. You can even manage distributivity that way.

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Additive Subgroup

Consider first the set $T = \{x - e;~x ∈ S\} = \{x - e;~x ∈ R,~ex = e = xe\}$. So for $y ∈ R$, $$y ∈ T \iff ey + e = e = ye + e.$$

Cancelling $e$, we have $T = \ker (e·~) ∩ \ker (~·e)$, so it’s an additive subgroup.

Now the addition you have defined is just the pullback of the addition under $$σ\colon S → T,~x ↦ x - e,$$ so $S$ is indeed a group with this addition.

Multiplicative Submonoid

Furthermore, $S ⊆ R$ is a multiplicative submonoid:

• $1 ∈ S$, as $1e = e = e1$ (as long as $R$ even has a unity $1$).
• $S·S ⊆ S$, as for $x,x' ∈ S$ we have $xx'e = xe = e = ex = exx'$.

Distributivity

Finally, to show that distributivity holds, it suffices to show that $σ$ commutes with all multiplications with arbitrary $s ∈ S$ from the left and right, as then, for all $x, x' ∈ S$, \begin{align*} \color{gray}{} s(x + x') = sx + sx' &\iff σ(s(x+x')) = σ(sx + sx')\\ & \iff s(σ(x + x')) = σ(sx) + σ(sx') \\ & \iff s(σ(x) + σ(x')) = sσ(x) + sσ(x'), \end{align*} with the same calculation going through for the right distributivity. But for $s ∈ S$ and $x ∈ S$, $$sσ(x) = s(x - e) = sx - se = sx - e = σ(sx),$$ with the same calculation going through for the right multiplication.

Ring Isomorphism $S → T$

The morphism $σ\colon S → T$ is even multiplicative, hence a ring isomorphism. For $x, x' ∈ S$, $$σ(x)σ(x') = (x - e)(x' - e) = xx' - xe - ex' + e = xx' - e = σ(xx'),$$ so $T$ becomes a subring of $R$ with $1 - e$ as unity (if $R$ has a unity $1$). Therefore, $σ$ becomes a ring isomorphism $S → T$.

Side note. This also suffices to show distributivity, because it says that $T$ is multiplicatively closed, so $T$ is a subring of $R$ and, in particular, distributivity holds for it. Because $σ$ is an additive und multiplicative bijection, distributivity must also hold in $S$.

$T$ As the Corner Ring

Let’s now assume that $R$ has a unity $1$.

Since $(1 - e)R ⊆ \ker (e· ~)$ as well as $R(1 - e) ⊆ \ker (~·e)$, we have $(1 - e)R(1 - e) ⊆ T$. On the other hand, as $(1 - e)$ is the identity on $T$, we have $$∀ t ∈ T\colon~ t = (1 - e)t(1 - e) ∈ (1 - e)R(1 - e)$$ and thus $T ⊆ (1 - e)R(1 - e)$. (Thanks to Darij Grinberg for pointing this out in the comments!)

All in all, $T = (1 - e)R(1 - e)$ and $σ$ is an isomorphism of $S$ and the corner ring $T$.