# Homework Help: Ring theory?

1. Dec 29, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Show that the intersection of any two subrings of a ring is a subring.

3. The attempt at a solution
It seems abstract.

suppose a+b=c and a*b=d

Then if c is in A and B (where A and B are subrings) then the intersection of A and B denoted by C contains c and if C contains more the one element then it must contain a and b.

My argument may not be general enough.

2. Dec 29, 2007

The argument doesn't seem to get to the point - you probably should use a "subring test theorem", here is one description on wiki: http://en.wikipedia.org/wiki/Subring_test

So take subrings A, B, and show the intersection satisfies the subring test.

3. Dec 29, 2007

### HallsofIvy

How does this follow? the way you have said it, c is just a member of C and has nothing to do with a and b. (There may be many different a and b such that a+ b= c for a given c.)
You need to show "if a and b are in C, then -a is in C, a+ b is in C, and a*b is in C. Since 0 and 1 must be in any ring, they must be in A and B and so in C.

4. Dec 29, 2007

### Mystic998

0 has to be in any ring, but a subring doesn't doesn't have to contain the multiplicative identity (assuming the original ring even has one). A good example is 2Z in Z.

Of course, that's entirely beside the point, but the question has already been answered, and I like being persnickety.

5. Dec 29, 2007

### Hurkyl

Staff Emeritus
I think the definition that rings have '1' is sufficiently pervasive that it should be assumed when not otherwise specified... honestly, the magma algebra system is the only context I've ever seen where a "ring" is not used to mean a unital associative algebra.

6. Dec 30, 2007

### pivoxa15

So the answer is quite simple?

7. Dec 30, 2007

### HallsofIvy

Yes, it is quite simple.

Let a and b be in the intersection of A and B, then a+ b is in A because ______

8. Dec 30, 2007

### ZioX

I still think the joke that goes 'I pronounce 'RNG' as wrong' is uproariously funny.