Algebra, Problems For Fun (36)

**NonCommAlg** · July 28th 2009, 11:04 PM

Definition: An element $r$ in a ring $R$ is called central if $r$ is in the center of $R,$ i.e. $rs=sr,$ for all $s \in R.$

1) Suppose $R$ is a ring in which $x^2=0 \Longrightarrow x=0.$ Let $e \in R$ be an idempotent, i.e. $e^2=e.$ Prove that $e$ is central.

Hint:

Spoiler:

2) Use 1) to give a short proof of this very special case of Jacobson's theorem: if $x^3=x,$ for all $x$ in a ring with identity $R,$ then $R$ is commutative.

Hint:

Spoiler:

3) This time suppose $x^4=x,$ for all $x$ in a ring with identity $R.$ Use 1) to prove that $R$ is commutative.

Hint:

Spoiler:

4) Challenge: Suppose $x^5=x,$ for all $x$ in a ring with identity $R.$ Use 1) to prove that $R$ is commutative.

**TheAbstractionist** · July 29th 2009, 06:16 AM

Originally Posted by NonCommAlg

Hint:

Spoiler:

Why is $r^2+r$ an idempotent? Isn’t $(r^2+r)^2=r^2+r+2r^3?$

**NonCommAlg** · July 29th 2009, 11:23 AM

Originally Posted by TheAbstractionist

Why is $r^2+r$ an idempotent? Isn’t $(r^2+r)^2=r^2+r+2r^3?$

because in $R$ we have $(-1)^4=-1,$ which gives us $2=0.$ so $2r^3=0.$

**TheAbstractionist** · July 29th 2009, 04:26 PM

Ah! I didn’t realize that $R$ was a ring with multiplicative identity.

**TheAbstractionist** · July 30th 2009, 09:38 AM

Right. For the last one, $r^4$ is idempotent and therefore central. Now

$(r+1)^5\,=\,r+1$

$\implies\ 2r^3+2r^2+r\,=\,-r^4$ is central

and

$(r-1)^5\,=\,r-1$

$\implies\ 2r^3-2r^2+r\,=\,r^4$ is central

and so $(2r^3+2r^2+r)-(2r^3-2r^2+r)=4r^2$ and $(2r^3+2r^2+r)+(2r^3-2r^2+r)=4r^3+2r$ are central. Hence $8(4r^2)=2^5r^2=2r^2$ is central. Moreover

$4r^3+6r^2+4r\,=\,(r+1)^4-r^4-1$

is central. Hence $(4r^3+6r^2+4r)-(4r^3+2r)-3(2r^2)=2r$ is central. Hence $(4r^3+2r)-2r=4r^3$ is central and so $8(4r^3)=2^5r^3=2r^3$ is central. Hence $(2r^3+2r^2+r)-2r^3-2r^2=r$ is central.

**NonCommAlg** · July 30th 2009, 11:46 AM

Originally Posted by TheAbstractionist

$(r+1)^5\,=\,r+1$

$\implies\ 2r^3+2r^2+r\,=\,-r^4$ is central

we can't divide by 5. so we actually get: $10r^3+10r^2+5r=-5r^4$ is central.

**TheAbstractionist** · July 31st 2009, 02:52 AM

Okay, let’s try again.

$10r^3+10r^2+5r=-5r^4$ and $10r^3-10r^2+5r=-5r^4<br />$ are central, so $(10r^3+10r^2+5r)-(10r^3-10r^2+5r)=20r^2$ and $(10r^3+10r^2+5r)+(10r^3-10r^2+5r)=20r^3+10r$ are central. So $2(20r^2)=(2^5+8)r^2=10r^2$ is central. Moreover

$4r^3+6r^2+4r\,=\,(r+1)^4-r^4-1$

is central. Hence $5(4r^3+6r^2+4r)-(20r^3+10r)-3(10r^2)=10r$ is central. Hence $(20r^3+10r)-10r=20r^3$ is central, and so $10r^3$ is central. Hence $(10r^3+10r^2+5r)-10r^3-10r^2=5r$ is central. Now

$4r^3-6r^2+4r\,=\,-(r+1)^4+r^4+1$

is also central, and so $(4r^3+6r^2+4r)-(4r^3-6r^2+4r)=12r^2$ is central, and so $12r^2-10r^2=2r^2$ is central. Replacing $r$ by $r+1$ gives that $2(r+1)^2$ is central. Hence $2(r+1)^2-2r^2-2=4r$ is central.

At last! Hence $r=5r-4r$ is central.

**NonCommAlg** · July 31st 2009, 06:10 PM

that's a nice work!

there are two points, which won't have any effect on your solution:

Originally Posted by TheAbstractionist

Hence $5(4r^3+6r^2+4r)-(20r^3+10r)-3(10r^2)=10r$ is central.

or since $30 = 0$ in R, we have $5(4r^3 + 6r^2 + 4r) - (20r^3 + 10r)=10r.$

$4r^3-6r^2+4r\,=\,-(r+1)^4+r^4+1$

the right hand side should be $-(r-1)^4+r^4+1.$

**aman_cc** · November 19th 2009, 11:33 PM

My question is related to:

2) Use 1) to give a short proof of this very special case of Jacobson's theorem: if

for all

in a ring with identity

then

is commutative.
Hint: Spoiler:
clearly

is idempotent, and thus central by 1), for any

Show that

and

are also central.

Let center of the ring be C. $C=\{r \in R | rs=sr \forall s \in R\}$
Clearly C is a sub-ring of R

Also $1 , r^2 \in C$ $\forall r \in R$
To prove $2r \in C$ $\forall r \in R$
Consider $(1+r)^2 - 1 - r^2 = 2r$
From the closure property of sub-ring we get $2r \in C$
Is my approach correct?

I am having trouble in proving $3r \in C$ . Any push plz?

**NonCommAlg** · November 19th 2009, 11:47 PM

Originally Posted by aman_cc

Let center of the ring be C. $C=\{r \in R | rs=sr \forall s \in R\}$
Clearly C is a sub-ring of R

Also $1 , r^2 \in C$ $\forall r \in R$
To prove $2r \in C$ $\forall r \in R$
Consider $(1+r)^2 - 1 - r^2 = 2r$
From the closure property of sub-ring we get $2r \in C$
Is my approach correct?

yes.

I am having trouble in proving $3r \in C$ . Any push plz?

$1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r.$ thus $3r=-3r^2 \in C.$

**aman_cc** · November 20th 2009, 08:51 AM

Originally Posted by NonCommAlg

yes.

$1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r.$ thus $3r=-3r^2 \in C.$

Thanks. It look so easy when you see it

I'm now stuck with
3) This time suppose

for all

in a ring with identity

Use 1) to prove that

is commutative.

Hint: Spoiler:
It's immediate that

is an idempotent, and thus central by 1), for all

Put

to show that

is central for all

As suggested in the hint - I could prove the following:

$\forall a,b \in R$
1. $a+a = 0$
2. $a^2 + a \in C$
3. $ab+ba \in C$

How do I show R is commutative? A push again plz?

Infact I was trying to show that ab+ba=0. Then using (1) above I will show
ab = ba and hence will be done. But no success.

**NonCommAlg** · November 20th 2009, 11:48 AM

Originally Posted by aman_cc

Thanks. It look so easy when you see it

I'm now stuck with
3) This time suppose

for all

in a ring with identity

Use 1) to prove that

is commutative.

Hint: Spoiler:
It's immediate that

is an idempotent, and thus central by 1), for all

Put

to show that

is central for all

As suggested in the hint - I could prove the following:

$\forall a,b \in R$
1. $a+a = 0$
2. $a^2 + a \in C$
3. $ab+ba \in C$

How do I show R is commutative? A push again plz?

Infact I was trying to show that ab+ba=0. Then using (1) above I will show
ab = ba and hence will be done. But no success.

so $a^2b + ba^2 \in C,$ for all $a,b \in R.$ thus $a^2(a^2b + ba^2)=(a^2b + ba^2)a^2,$ which gives us: $ab=ba.$

**aman_cc** · November 21st 2009, 04:04 AM

Originally Posted by NonCommAlg

so $a^2b + ba^2 \in C,$ for all $a,b \in R.$ thus $a^2(a^2b + ba^2)=(a^2b + ba^2)a^2,$ which gives us: $ab=ba.$

Thanks. And now for the last part.

4) Challenge: Suppose

for all

in a ring with identity

Use 1) to prove that

is commutative.

I completely follow TheAbstractionist proof. I have just two question.

1. I would never have been able to do this proof. Is there a structured line of attack which you were following here? Or this just a matter of lot of practice and off-course lot of gray-matter?

2. Do we end at 5? Or this is true in general, where $x^n=x$ ?

**NonCommAlg** · November 21st 2009, 04:06 PM

Originally Posted by aman_cc

Thanks. And now for the last part.

4) Challenge: Suppose

for all

in a ring with identity

Use 1) to prove that

is commutative.

I completely follow TheAbstractionist proof. I have just two question.

1. I would never have been able to do this proof. Is there a structured line of attack which you were following here? Or this just a matter of lot of practice and off-course lot of gray-matter?

the idea is to look for idempotents.

2. Do we end at 5? Or this is true in general, where $x^n=x$ ?

the result is true for all $n.$ the interesting thing is that $n$ can even change as the element $x$ changes. so, the Jacobson's theorem says:

if $R$ is a ring with identity such that for every $x \in R,$ there exists an integer $n(x) \geq 2$ such that $x^{n(x)}=x,$ then $R$ is commutative.

of course, understanding the proof of this theorem requires a fairly deep knowledge of ring theory. there are, however, special cases of

this theorem which can be proved quite easily. for example if $R$ is finite or, more genrally, Artinian, then the theorem is just a quick result

of Artin-Wedderburn theorem.

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