The ideals of Zn are, first of all, additive subgroups of Zn. These we know to all have the form 〈d〉 where d divides n. But, as we know, the set 〈d〉 is the ideal generated by d. So we have just proven that The ideals in Zn are precisely the sets of the form 〈d〉 where d divides n.
How do you find the Z12 ideals?
The ring Z12 is commutative, we have x·y = y·x ∈ 〈4〉, for all x ∈ 〈4〉, and all y ∈ Z12. Therefore 〈4〉 is an ideal of Z12.
What are the maximal ideals of Zn?
By this, the maximal ideals in Zn are in bijection with the maximal ideals of Z containing Ker(φ) = nZ. The maximal ideals of Z are all of the form (p) for primes p, and it is easily checked that such an ideal contains (n) if and only if p | n.
How many ideals are in Z8?
The positive divisors of 8 are 1, 2, 4 and 8, so the ideals in Z8 are: (1) = Z8, (2) = {0, 2, 4, 6}, (4) = {0, 4}, (8) = {0}. Of these, by inspection (2) is maximal (and therefore prime), whereas (1) and (8) are improper, so neither prime nor maximal.
Is every ideal of Zn a principal ideal?
One of the important properties of Z, not held by rings in general, is that every ideal of Z is principal. Theorem 1.17. Let a ✓ Z be an ideal. Then there is an m 2 Z such that a = (m) = mZ.
What are the prime ideals of Z12?
For R = Z12, two maximal ideals are M1 = {0,2,4,6,8,10} and M2 = {0,3,6,9}. Two other ideals which are not maximal are {0,4,8} and {0,6}.
How do you find the number of ideals?
Since R is local artinian we must have Ki=m=maximal ideal of R. This gives m=Ann(ai). This means that R contains the field k=R/m. This means that R is a finite dimensional vectorspace.
How do you find maximal ideals?
Given a ring R and a proper ideal I of R (that is I ≠ R), I is a maximal ideal of R if any of the following equivalent conditions hold: There exists no other proper ideal J of R so that I ⊊ J. For any ideal J with I ⊆ J, either J = I or J = R. The quotient ring R/I is a simple ring.