Question

The number of proper ideals of Z17 is​

Answer 1

Step-by-step explanation:

Any ideal of a ring which is strictly smaller than the whole ring. For example,  is a proper ideal of the ring of integers , since .

The ideal  of the polynomial ring  is also proper, since it consists of all multiples of , and the constant polynomial 1 is certainly not among them.

In general, an ideal  of a unit ring  is proper iff The latter condition is obviously sufficient, but it is also necessary, because would imply that for all 

so that a contradiction.

Note that the above condition follows by definition: an ideal is always closed under multiplication by any element of the ring. The same property implies that an ideal containing an invertible element cannot be proper, because , where denotes the multiplicative inverse of in

Since in field all nonzero elements are invertible, it follows that the only proper ideal of  is the zero ideal.

Answer 2

The number of proper ideals of Z17 will be 0.

Given,

Z17 = (17 elements) = (0,1,2,3,....,17)

To Find,

The number of Proper Ideals of Z17.

Solution,

Z17 = (17 elements) = (0,1,2,3,....,17)

We know that 17 is divisible only by 1 and 17. So there will be two ideals. They are given below:

For Ideal 1,

$\frac{1}{17} = < 1 > = {0,1,2,3,..17}$

For Ideal 2,

$\frac{17}{17} = < 17 > = { 0 }$

For Ideal 1,

Since maximal Ideal for Z17 cannot be equal to all the elements under Z17,

Therefore Ideal 1 is considered as an improper ideal and not the maximal or proper one.

Now for Ideal 2,

only element contained by Ideal 3 is 0, therefore we cannot consider it as a proper ideal. It will be Trivial Ideal.

Hence, the number of proper ideals of Z17 will be 0.

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