# Ideal Rings - Abstract Algebra

1. Apr 3, 2009

### golfgreen99

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

Suppose R is a ring and I,J is an ideal to R.

Show (i) I+J is ideal to R. (ii) I union J is ideal to R.

2. Relevant equations

none

2. Apr 3, 2009

### foxjwill

Can you show that you've attempted it?

3. Apr 3, 2009

### golfgreen99

Here is what I have (I think I have (ii)):

(i)
Part 1: Clearly if a,b are elements of I, then a+b are elements of I (since I is ideal to R) and if a,b are elements of J then a + b are elements of J (since J is ideal to R)

If we have one element from each ideal, say a element of I/b element of J, then we must show a+b is an element of (I+J).

a+b is an element of R since a,b is an element of R and R is closed.

Part 2: Let a be an element of (I + J) and r is an element of R.

Then if a is an element of I, ar is an element of I and ra is an element of I since I is ideal.

Similarly, if a is an element of J, ar is an element of J and ra is an element of J since I is ideal.

In either case, ar is an element of (I + J) and ra is an element of (I + J)

(ii)
Part 1: Assume a, b are elements in I union J.

Then a + b is an element of I since I is ideal and a + b is an element of J since J is ideal.

Then a + b is in both I and J and therefore a + b is an element of I union J.

Part 2: Assume a is and element of (I union J) and r is an element of R.

Then, ar is an element of I and ra is and element of I since I is ideal.

Similarly, ar is an element of J and ra is an element of J since J is ideal.

So, ar is an element of (I union J) and ra is an element of (I union J)

4. Apr 3, 2009

### foxjwill

you'll also need to show that I+J and $$I\cup J$$ are closed under multiplication and subtraction (in fact addition follows from subtraction).

For (i) part 1, it might help to recall that since $$(I,+,\cdot)$$ is an ideal of $$(R,+,\cdot)$$, $$(I,+)$$ is a normal subgroup of $$(R,+)$$ (since $$(R,+)$$ is abelian). Simlarly for J.