# Abstract Algebra - Show that A+B and AB are Ideals in R?

Oh this is fun one *:) , I'm just a little stuck on how to get started, so I'd appreciate any help :)

If there are 2 ideals, A and B of a ring R, and their sum and product are as follows:

A + B = {a +b | a ∈ A and b ∈ B} and,

AB = {a_1*b_1 +a_2*b_2+ ... + a_n*b_n | n > 0 , and a_i ∈ A and b_i ∈ B for 1 ≤ i ≤ n}.

How do I show that A +B and AB are also ideals of R?

This really seems like fun question, so any hint you guys have would be really appreciated for me to get this one done.

Thanks,

M. :)

### 1 Answer

- DavidLv 78 years agoFavorite Answer
first we assume that r,r' are in A+B. then we check if r-r' is also in A+B. this will show that A+B is an additive subgroup of R.

from our initial assumption, we may write r = a+b, for some a in A, b in B, and likewise r' = a'+b'.

therefore: r - r' = (a+b) - (a'+b') = (a-a') + (b-b'), which is in A+B.

now suppose x is any element of R, and r is in A+B. we must show that xr is in A+B.

again writing r = a+b, we have xr = x(a+b) = xa + xb.

since A is an ideal, xa is in A, since B is an ideal, xb is in B.

thus xr is in A+B. (technically, this only shows A+B is a left ideal

if A and B are ideals. to prove A+B is a two-sided ideal, one must also show

that rx is also in A+B, the proof is similar (and uses the fact that

A and B are two-sided ideals). if our ring is commutative, this is unnecessary).

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showing AB is an ideal is a bit more involved:

let r = a1b1 + a2b2 +...+ anbn, and r' = a'1b'1 + a'2b'2+...a'mb'm

then r - r' is an element of AB with m+n "terms"

(since -a'jb'j = (-a'j)(b'j), and -a'j is in A iff a'j is), so r - r' is in AB.

that AB is a left ideal depends only on A being a left ideal:

x(Σ ajbj) = Σ x(ajbj) = Σ (xaj)(bj), that AB is a right ideal

depends only on B being a right ideal.

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