# Thread: order of ring and idempotent

1. ## order of ring and idempotent

Q1) Consider the matrix ring M2(Z2)
a) Find the order of the ring
b) List all the units of the ring

Q3) an element a in a ring is an idempotent if a^2=a
show the set of idempotents in a commutative ring are closed under multiplication and find the idempotents in
Z8 X Z12

2. Originally Posted by Ryan0710
Q1) Consider the matrix ring M2(Z2)
a) Find the order of the ring
b) List all the units of the ring
a) EDIT: I was wrong. The order as seen below is 8. I apologise for my mistake. Hey Ryan, how come you accepted my answer when it was not convincing?

b)For the multiplicative inverse to exist , the determinant must be non-zero. Now since the determinant for $\displaystyle \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$ is $\displaystyle ad-bc$. And since $\displaystyle a,b,c,d \in \mathbb{Z}_2$, either ad = 1 or bc = 1 is the only possible solution.

Thus the only units are:

$\displaystyle \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right),\left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right),\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)$

With bc =1,
$\displaystyle \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right),\left(\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}\right),\left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right)$

3. Originally Posted by Ryan0710
Q3) an element a in a ring is an idempotent if a^2=a
show the set of idempotents in a commutative ring are closed under multiplication and find the idempotents in
Z8 X Z12
let $\displaystyle E \subset R$ and E is the set of idempotent elements of R.

if $\displaystyle x,y \in E$ then $\displaystyle x^2,y^2 \in E$

$\displaystyle x\cdot y=x^2\cdot y^2=(x\cdot x)\cdot(y \cdot y)$

Using both the associtative and communitive properties we get

$\displaystyle =(x\cdot y)\cdot(x \cdot y)=(xy)\cdot(xy)=(xy)^2 \in E$

Therefore E is closed

Edit:Sorry about the mistake. Thanks NonCommAlg.

4. 2 little points:

1) the order of a ring is the number of elements of that ring. thus the order of $\displaystyle M_2(\mathbb{Z}_2)$ is 8.

2) in $\displaystyle \mathbb{Z}_8$ only 0 and 1 are idempotent. the idempotents of $\displaystyle \mathbb{Z}_{12}$ are 0, 1, 4 and 9. so there are

8 idempotents in $\displaystyle \mathbb{Z}_8 \times \mathbb{Z}_{12}.$ what are they Ryan0710 ?

5. Originally Posted by NonCommAlg
2 little points:

1) the order of a ring is the number of elements of that ring. thus the order of $\displaystyle M_2(\mathbb{Z}_2)$ is 8.
I am sorry, I am not a student of algebra in a formal school. I just read this. It claimed the order of the ring is the order of the additive group. Isnt the order of the additive group 2?

To Ryan,

I must be wrong, NonCommAlg is a good algebraist, there is no way he would have been wrong on such "elementary" things

I apologise for the mistake.

6. Originally Posted by Isomorphism
I am sorry, I am not a student of algebra in a formal school. I just read this. It claimed the order of the ring is the order of the additive group. Isnt the order of the additive group 2?

To Ryan,

I must be wrong, NonCommAlg is a good algebraist, there is no way he would have been wrong on such "elementary" things

I apologise for the mistake.
the order of a (finite) group is the number of elements of that group. by the additive group of a ring R, we mean the ring R itself

considered as an (abelian) additive group, i.e. (R, +). so the set is still the ring R, but considered with + only. so the order of a (finite)

ring is the number of elements of that ring. regarding your question, 2 is the order of every (non-zero) element of $\displaystyle M_2(\mathbb{Z}_2),$ and not

the order of the ring.

7. Originally Posted by NonCommAlg
the order of a (finite) group is the number of elements of that group. by the additive group of a ring R, we mean the ring R itself

considered as an (abelian) additive group, i.e. (R, +). so the set is still the ring R, but considered with + only. so the order of a (finite)

ring is the number of elements of that ring. regarding your question, 2 is the order of every (non-zero) element of $\displaystyle M_2(\mathbb{Z}_2),$ and not

the order of the ring.
Oh!

I am sorry. I confused the order of the group with the order of an element.

Thank you

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