# Thread: Prove that A/I has the same characteristic as A.

1. ## Prove that A/I has the same characteristic as A.

(a) Suppose that $A$ is an integral domain of positive characteristic. Suppose that $I \subsetneq A$ is an ideal. Prove that $A/I$ has the same characteristic as $A$.

(b) Give an example which demonstrates that $A/I$ need not be an integral domain.

2. Originally Posted by JJMC89
(a) Suppose that $A$ is an integral domain of positive characteristic. Suppose that $I \subsetneq A$ is an ideal. Prove that $A/I$ has the same characteristic as $A$.

Hints - Try to show the following:

1) The characteristic of any integral domain is either zero or a prime

2) If for an abelian group G we define $exponent(G):=Exp(G):=\max\{n\in\mathbb{N} | n=ord(x)\mbox{ for some }x\in G\}$ , then

the exponent of any subgroup of G is a divisor of the group's exponent or zero if the

latter is zero.

(b) Give an example which demonstrates that $A/I$ need not be an integral domain.

Take $A:=\mathbb{F}_p[x]\,,\,\,I=$ , with $\mathbb{F}_p$ = the prime field

of characteristic p.

Tonio

3. 1) I did this as a previous proof.

2) What does the exponent have to do with proving that the quotient ring has the same characteristic as the integral domain?

4. Originally Posted by JJMC89
1) I did this as a previous proof.

2) What does the exponent have to do with proving that the quotient ring has the same characteristic as the integral domain?

An ideal is also an additive subgroup of the abelian additive group of the ring, and then

in the quotient group $A/I$ ...

Tonio

5. So you're saying that the exponent of the Abelian additive subgroup of the ring and the characteristic of the ring are the same and that the exponent of the quotient group is the same as the characteristic of the quotient ring?

6. Originally Posted by JJMC89
So you're saying that the exponent of the Abelian additive subgroup of the ring and the characteristic of the ring are the same and that the exponent of the quotient group is the same as the characteristic of the quotient ring?

I'm saying the former (it's the definition of ring characteristic!), and the latter is what you have to prove...

Suppose $p>0$ is the characteristic (exponent) of $R$ , and take $1+I\in R/I$ .

Clearly $p(1+I)=p\cdot 1+I=0\mbox{ in } R/I\Longrightarrow$ the exponent of the

quotient ring has to be a divisor of $p$, but since this is a prime...

Tonio

7. Originally Posted by tonio
it's the definition of ring characteristic!
The definition I was given for ring characteristic is different, but now I see that they are the same.

I've got it now. Thanks.