Prime Ideals and Functions

gistfile1.txt

Claim: there's a bijection between prime ideals of a commutative ring R

(I is a prime ideal of R iff:
 (1) (I, +) is a subgroup of (R, +)
 (2) if r ∈ R and x ∈ I, then rx ∈ I
 (3) if ab ∈ I, then a ∈ I or b ∈ I
 (4) 1 ∉ I)

and maps f : R → 2 s.t.
(a) f(0) = 0
(b) f(1) = 1 and f(-1) = 1
(c) f(a * b) = min(f(a), f(b))
(d) f(a + b) ≤ max(f(a), f(b))

Proof sketch:

### Direction from ideal to function.

If we have a prime ideal I, convert it to a function f by
setting
f(x) = x ∈ I ? 0 : 1
I ⊆ R means f(0) = 0 and f(a) = 0 and f(b) = 0 imply f(a + b) = 0
1 ∉ I means f(1) = 1
Also if I is a subgroup, we have to have f(-1) = 1, because if f(-1) = 0,
then f(1) = 0, a contradiction.
So we have (a) and (b)
(3) means that if f(a) = 1 and f(b) = 1, then f(ab) = 1.
(2) means that if f(a) = 0, then f(ab) = 0.
So combining (2) and (3), we have (c).
All that remains is (d).
Consider all possibilities for f(a) and f(b).
If f(a) = f(b) = 0, we must show f(a + b) = 0. But this follows from (1).
Otherwise max(f(a), f(b)) = 1, and f(a + b) ≤ 1 is trivially true.

### Direction from function to ideal 

If we have a function f : R → 2, then I is defined by
I = {x ∈ R | f(x) = 0}
We have f(0) = 0. We have that if f(a) = 0 and f(b) = 0, then
f(a + b) = 0, by (d). Suppose f(a) = 0. Is it the case that f(-a) = 0?
Well, f(-a) = f(-1 * a) = min(f(-1), f(a)) = min(1, f(a)) = f(a) = 0,
so yes.
So we have (1).
For (2), we must show If r ∈ R and f(x) = 0, then f(xr) = 0. But this follows from (c).
For (3), we must show that if f(ab) = 0, then f(a) = 0 or f(b) = 0. But this is
the contrapositive of f(a) = 1 and f(b) = 1 implies f(ab) = 1, an instance of (c).
Finally, (4) follows from (b).

Ugly but true :-)

Agreed, I hope it points the way to something cleaner :)