Ring theory?

Question

How would you show that A = {3x,y l x,y are integers} is a maximal ideal of Z + Z where + refers to direct product.???

Answer 1

With any maximal subgroup or subring problem you need to assume not and suppose that there is a subgroup/subring between the given one and the whole ring/group. In this example, you will need to use the fact that 3 is prime.

Answer 2

First, show it is an ideal:
- Show (A, +) is a subgroup of (Z + Z, +):
- (3a, b) + (3c, d) = (3(a+c), b+d), so it is closed
- (3a, b) + (3(-a), -b) = (3a + (-3a), b+-b) = (0, 0), so inverses exist.
- Show that for any a ∈ A, any r ∈ Z+Z, ra and ar are in A.
- Let a = (3x, y) ∈ A and r = (c, d) ∈ Z+Z.
Then ra = ar = (3xc, yd) = (3(xc), yd) ∈ A.

Now, to show it is a maximal ideal, suppose I is an ideal of Z+Z strictly containing A and let (x, y) ∈ I - A. We will show that I must contain (1, 0) and thence that I must be equal to Z+Z. Write x = 3k + r, where r must be 1 or 2. Since A and hence I contains (3(-k), y), I must also contain (x, y) + (-3k, y) = (r, 0).

If r = 1 then I obviously contains (1, 0). If r = 2 then I contains (2, 0) and (-3, 0) ∈ A, so I contains (2, 0) + (2, 0) + (-3, 0) = (1, 0). So in either case I contains (1, 0). [Note that this step is the one that relies on 3 being prime. It's obscured here since I've simply listed cases, but if the number used is not prime we can't get (1, 0) in all cases and the rest of the proof does not work.]

I also contains (0, 1) ∈ A. But (1, 0) and (0, 1) generate Z+Z, so I = Z+Z. So there is no proper ideal of Z+Z strictly containing A. Hence A is a maximal ideal of Z+Z.

Answer 3

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Answer 4

you can do as above or use the propositions:
1)M is maximal ideal iff R/M is a field
2) If I_1 is ideal in R_1 and I_2 in R_2 then
(R_1 + R_2)/(I_1 + I_2) isomorphic with R_1/I_1 + R_2/I_2

Now A is actually 3Z + Z, where + is the direct product
then (Z + Z)/(3Z + Z) isomorphic with Z/3Z + Z/Z = Z/3Z since Z/Z = 0.
But Z/3Z is a field which implies that 3Z + Z is maximal ideal.
Straightforward!