Suppose H is a subring of R such that for every $r∈H$ we have $r^2=e$. Prove that $H$ is commutative.

Question

So I know that since H is a subring of $\mathbb{R}$ that H is then by definition:

1) closed under addition

2) closed under multiplication

3) have the same multiplicative identity as $\mathbb{R}$

Am I missing anything? I think I need to only reference fact 2, which states that H is closed under multiplication for this proof.

I tried looking up on the internet how to show that a subring is commutative and did not find anything. Is the answer as simple as saying that since $\mathbb{R}$ is commutative then the subring H, is itself by definition commutative?

Or do I go about it by showing that lets say for example:

Let $r_1, r_2 \in H$ then $r_1 \times r_2 = e = r_1 \times r_2$ which is then commutative?

It seems that you are writing $e$ for the multiplicative identity of the ring, which is usually written $1$. Under your assumptions, $0 = 0^2 = 1$, which implies that $H$ and $R$ are both the trivial ring containing just one element. The trivial ring is commutative. — Rob Arthan, Mar 29 '20 at 20:34
Why do you use $\mathbb{R}$ as notation for ring? Very confusing! — J. De Ro, Mar 29 '20 at 20:50
It is highly unlikely the problem meant to trivialize everything by asking about a subring of a commutative ring. — rschwieb, Mar 29 '20 at 21:23
If every nonzero element squares to $e$, clearly every nonzero element is a unit, and the set of nonzero elements is a group. That makes it fall under a frequently asked group theory question. — rschwieb, Mar 30 '20 at 14:22

score 4 · Answer 1 · answered Mar 29 '20 at 19:58

4

Commutativity means that for all $r,s\in H$ we have that $r\circ s= s\circ r$ where $\circ: H\times H\to H$ denotes the multiplication of $R$ restricted to $H$.

Now to address your question: If for all $r\in H$ we have that $r\circ r=e$, then, $r=r^{-1}$ holds for all $r\in H$. Thus, given $r,s\in H$, $$r\circ s = r^{-1}\circ s^{-1}=(s\circ r)^{-1}=s\circ r,$$ where for the last equality we used that $s\circ r\in H$.

answered Mar 29 '20 at 19:58

sampleuser

419

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So what about the fact that the conditions were supposed to be for a subring and that they cannot hold then? – Tobias Kildetoft Mar 29 '20 at 20:34
@TobiasKildetoft: I think the conditions can hold, but they show that $H$ and $R$ are trivial. See my comment on the question. – Rob Arthan Mar 29 '20 at 20:36
@Rob My guess would be that the OP forgot some part of the question. But in either case, it calls for a clarifying comment like yours, rather than an answer like this one that misses some important details. – Tobias Kildetoft Mar 29 '20 at 20:38
@TobiasKildetoft: absolutely! – Rob Arthan Mar 29 '20 at 20:40
What I have in the question is directly copy and pasted from the homework – BlackKnightRider Mar 29 '20 at 20:41
Does this not follow then? – BlackKnightRider Mar 29 '20 at 20:41
@Tobias Kildetoft I don’t understand what you want to point at. These conditions can hold and the argument is perfectly fine. – sampleuser Mar 29 '20 at 20:52
@sampleuser: what is $0^{-1}$ in an arbitrary ring? – Rob Arthan Mar 29 '20 at 20:53
As you pointed out already, the assumptions imply that R is the trivial ring with 0=1. thus $0^{-1}=1$. – sampleuser Mar 29 '20 at 20:54

Robert Lewis · Accepted Answer · 2020-03-29T20:47:50.707

4

Let us assume that every $0 \ne r \in H$ satisfies $r^2 = e$, ruling out the case $e = 0^2 = 0$, which trivializes everything. These words inspired by the comment of Rob Arthan to the question itself.

Let

$x, y \in H; \tag 0$

then

$x^2 = y^2 = e; \tag 1$

also,

$xy \in H, \tag{1.5}$

whence

$xyxy = (xy)^2 = e; \tag 2$

then

$yxy = eyxy = x^2yxy = x(xyxy) = xe = x, \tag 3$

and

$yx = yxe = yxy^2 = (yxy)y = xy, \tag 4$

which shows $H$ is commutative. $OE\Delta$.

edited Mar 29 '20 at 20:47

answered Mar 29 '20 at 20:20

Robert Lewis

72,871

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So what about the fact that the conditions were supposed to be for a subring and that they cannot hold then? – Tobias Kildetoft Mar 29 '20 at 20:34
@TobiasKildetoft: nor sure I under.stand your question – Robert Lewis Mar 29 '20 at 20:44
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your logic looks perfectly fine to me. I think he thinks I left some details out in the question but I posted it exactly as is – BlackKnightRider Mar 29 '20 at 20:46
Please stop the practice of duplicating solutions over and over. I mean, you have written exactly this answer before, and also here and not in detail here . I won't speculate on why you do this, but I will suggest to you (not for the first time, I believe) to prefer duplication above diluting the answer pool with yet-another copy of the same thing. – rschwieb Mar 30 '20 at 14:32

score 2 · Answer 3 · answered Mar 29 '20 at 20:50

2

I assume you are using $e$ to denote the multiplicative identity of the ring $R$. I will follow custom and write $1$ rather than $e$. If $H$ is a subring of a ring $R$, then the additive identity $0$ is a member of $H$. As $0^2 = 0$, if $0^2 = 1$ we have $0 = 1$ which implies $x = 1x = 0x = 0$ for every $x \in R$. Thus your hypothesis implies that $R$ (and $H$) is the trivial ring with just one element $0 = 1$. The trivial ring is commutative.

answered Mar 29 '20 at 20:50

Rob Arthan

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Yes I am using e to identify the multiplicative identity, Okay I will add that logic to the opening. I assume my professor wants to see both cases – BlackKnightRider Mar 29 '20 at 20:53

Suppose H is a subring of R such that for every $r∈H$ we have $r^2=e$. Prove that $H$ is commutative.

3 Answers3