Question

Number of maximal ideals of Z-36, is
(A)3
(B) 2
(C)4
(D) None of these​

Answer 1

Answer:

In the ring Z of integers, the maximal ideals are the principal ideals generated by a prime number. More generally, all nonzero prime ideals are maximal in a principal ideal domain.

Answer 2

Answer:

There are four maximal ideals of Z/36Z, and the answer is (C) 4.

Step-by-step explanation:

The ring Z/36Z is a finite commutative ring with identity. We know that if R is a commutative ring with identity and m is a maximal ideal of R, then R/m is a field.

In this case, Z/36Z is the ring of integers modulo 36. The ideals of Z/36Z are precisely the sets of residue classes that form subrings of Z/36Z. The maximal ideals of Z/36Z are the maximal elements among these subrings, with respect to inclusion.

Note that Z/36Z is not a field, since 36 is not a prime number. Thus, there must exist at least one maximal ideal of Z/36Z.

To find the maximal ideals of Z/36Z, we can use the fact that Z/36Z is isomorphic to the product of the rings Z/4Z and Z/9Z (by the Chinese Remainder Theorem). The ideals of Z/4Z are precisely {0}, {1}, {2}, and {3}, and the ideals of Z/9Z are precisely {0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, and {8}.

The ideals of Z/36Z are then the sets of residue classes that can be written as a product of an ideal of Z/4Z and an ideal of Z/9Z. The maximal ideals of Z/36Z are the maximal elements among these sets, with respect to inclusion.

To find the maximal ideals, we can simply list out all possible products of maximal ideals of Z/4Z and Z/9Z. There are four maximal ideals of Z/4Z ({2Z/4Z}, {1+2Z/4Z}, {2+2Z/4Z}, and {3+2Z/4Z}), and nine maximal ideals of Z/9Z ({3Z/9Z}, {6Z/9Z}, {2+3Z/9Z}, {5+3Z/9Z}, {1+6Z/9Z}, {4+6Z/9Z}, {2+7Z/9Z}, {5+7Z/9Z}, and {8+7Z/9Z}).

Therefore, there are 4x9=36 possible products of maximal ideals of Z/4Z and Z/9Z. However, we need to eliminate the products that are not maximal in Z/36Z.

For example, {2Z/4Z}x{3Z/9Z}={6Z/36Z} is not maximal in Z/36Z, since it is properly contained in the ideal {0, 6, 12, 18, 24, 30}.

After eliminating the products that are not maximal, we are left with the following maximal ideals of Z/36Z:

{0}x{3Z/9Z}={0+9Z/36Z}

{0}x{6Z/9Z}={0+18Z/36Z}

{2+2Z/4Z}x{3Z/9Z}={6+18Z/36Z}

{2+2Z/4Z}x{6Z/9Z}={6+27Z/36Z}

Therefore, there are four maximal ideals of Z/36Z, and the answer is (C) 4.

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