Showing a subset is a subring?

1. Apr 9, 2012

SMA_01

Showing a subset is a subring???

1. The problem statement, all variables and given/known data

Let R be a ring and a a fixed element in R. Let Ia={x in R l ax=0}

2. Relevant equations

3. The attempt at a solution

I saw these conditions in my book, but I'm not sure are these conditions sufficient in showing Ia is a subring?

(1) 0 is in Ia:

Let 0 be in R, then
a(0)=0

(2) (a-b)is in Ia, for a, b in Ia:

I'm not sure how I should start this.

(3) Ia is closed under multipication.

2. Apr 9, 2012

kru_

Re: Showing a subset is a subring???

For 2) Let m, n be any two elements from Ia. Can you show that m+n is still in Ia? What do m and n look like? What does their sum look like?

It is similar for showing that m*n is in Ia, as well.

3. Apr 9, 2012

SMA_01

Re: Showing a subset is a subring???

Okay, so for 2 this is what I got:
Let m,n be in Ia. Then this means that am=0 and an=0. So,
am=an
and am-an=0
so, a(m-n)=0 therefore (m-n) is in Ia. Is this correct?
Should I follow the same flow for 3?

Thanks

4. Apr 9, 2012

kru_

Re: Showing a subset is a subring???

Looks good.

5. Apr 9, 2012

kru_

Re: Showing a subset is a subring???

Note, you also need to show the distributive property of multiplication over addition holds, and associativity of multiplication and addition.

6. Apr 10, 2012

SMA_01

Re: Showing a subset is a subring???

Thank you

7. Apr 10, 2012

micromass

Staff Emeritus
Re: Showing a subset is a subring???

No, you do not have to show those. Just showing the three conditions in the OP is good enough.

8. Apr 10, 2012

kru_

Re: Showing a subset is a subring???

Haha. Not according to my old abstract professor! We know they are inherited by R, but failure to show is -2 points..

sigh..