Let R=Z27. Find a. N(⟨0⟩). b. N(⟨3⟩). c. N(⟨9⟩).

Name: Problem 49, Chapter 14: Contemporary Abstract Algebra
Uploaded: 2021-07-18T19:47:44-08:00
Duration: 7 min 10 s
Channel: Gaurav Kalra
Description: Let R=Z27. Find a. N(⟨0⟩). b. N(⟨3⟩). c. N(⟨9⟩).

Key Concepts

Norm of an Ideal

In a finite commutative ring, the norm of an ideal I is defined as the number of elements in the quotient ring R/I. This concept is an analogue of the norm of an ideal in algebraic number theory, where it measures the “index” of the ideal in the ring. When R is finite, finding the norm of I is simply a matter of computing |R/I|.

Ideals in Z_n

The ring Z_n has a very simple ideal structure. Every ideal in Z_n is of the form ?d? where d divides n. In Z_27, for example, the nonzero proper ideals correspond to the divisors of 27. This structure allows us to effectively determine both the order of the ideal (as a subgroup of the additive group) and the index, which is used in computing the norm.

Quotient Rings and Their Cardinality

Given an ideal I in a ring R, the quotient ring R/I is the set of cosets of I in R. Its cardinality is equal to the index of I in R, meaning it is the number of distinct cosets. In the case of a finite ring, if |R| and |I| are known, then |R/I| = |R|/|I|. This is a direct application of Lagrange’s theorem in the context of finite groups, when viewing a ring’s additive group.

Index of a Subgroup in Z_n

In the additive group of Z_n, any ideal is also a subgroup, and its index is the ratio of the size of Z_n to the size of the subgroup. Determining the index in this way allows us to compute the norm of an ideal, since for an ideal I, the norm N(I) is given by the index of I in Z_n.

Transcript

00:02 I have been giving here snr as c0 negative c1 plus c2 and so on for s and r as 2n as 28.

00:12 We'll check for the values of nnr for which it is correct.

00:16 Similarly for negative 15 and then we find out by putting rs n by putting rs and plus 1.

00:24 So let's look at here the series is c not negative c1 plus c2, negative c3 and so on.

00:31 So if we just write out here two series, is the first is given as we have 1 plus x plus x squared and so multiply with second series we write the first term here we have c north right so here we have it is one so we write here c not now next year we have it is negative c1 so right as negative c1 x plus we have c2 x here, negative, and so on it is.

01:11 So, negative 1 to the power n, pn extra part n.

01:17 If you look at here at this c, 0 negative c1, x plus c2, x per and so on, plus negative 1 to the power n, cn, x.

01:27 This it is, is an expansion of 1 negative x to the power n.

01:37 But now we're to find out this sum here.

01:40 T0 negative c1 plus c2, negative c3, and so on.

01:46 So this sum we need to find out here.

01:50 So if you just look at here, this series is multiplying here with this series.

01:55 When this is given as 1 for the xxxxx0 and so on, it has given as 1 negative x to the power negative 1.

02:03 And we just multiply here this series and this series.

02:08 We multiply this with this.

02:09 So let's analyze for this whole series here product, the coefficient of x2 power r.

02:17 So for a proportion of x2 power r if we analyze, so if in this product here, let me take here, for example, r as 1, so the coefficient of x2 power r that is given as c node.

02:30 If i take r s2, then i will get it as actually we start with zero here.

02:39 But for r s0, we get it as c0.

02:42 Then for r as 1, if you take here now, r as 1, so we have x2 power 1.

02:51 And see here x2 power 1, this term times this term.

02:55 We get negative c1, and then x times c0.

02:59 We get c0.

02:59 So we get it as c0 negative c1.

03:02 You can see here the term.

03:04 That is c0 negative c1.

03:07 Similarly, for r equals 2, you will get c0 negative c1 plus c2.

03:10 At the coefficient of x2, we can say, if you take the product of this, the coefficient of x2 power r will give this term as nr so here we have the two cds here so we just multiply them one negative x to the power negative 1 1 negative x to the power n is coming out to be 1 negative x to the power n negative 1 i'm going to find out the coefficient for x2 part r that is given as negative 1 with the power r and negative 1 c r now the sum is given as the first part is sum is given as s and r equals 28.

03:51 If we put it equals to 28.

03:55 And now we check the values here for let's say n equals 9 r equals 2 r equals 2.

04:04 We take that on the left -hand side.

04:06 So we get lhs equal negative 1 to the part 2...

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