Showing a subset is a subring?

[SMA_01]

Showing a subset is a subring?

Homework Statement

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

Homework Equations

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.

 

[kru_]

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.

 

[SMA_01]

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

 

[kru_]

Looks good.

 

[kru_]

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

 

[SMA_01]

Thank you

 

[micromass]

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

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

 

[kru_]

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

sigh..