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Why Read This Book? - Index of

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340 Chapter 9 Rings<br />

will show in the next exercise. True to form, you suppose I + a = I + b and<br />

I + c = I + d and use this to show that I + ac = I + bd. That is, supposing a − b<br />

and c − d are in I should somehow allow you to show that ac − bd is in I also.<br />

EXERCISE 9.15.3 Finish the pro<strong>of</strong> <strong>of</strong> Theorem 9.15.2 by showing that multiplication<br />

is well defined.<br />

With R/I defined and shown to be a ring, we should be able to bypass all<br />

the chatty exposition as in Chapter 8 and jump right to results analogous to<br />

Theorems 8.6.19 and 8.6.21.<br />

EXERCISE 9.15.4<br />

(a) State a result analogous to Theorem 8.6.19 for a ring and the quotient ring<br />

created by “modding” out an ideal.<br />

(b) Explain how all except one component <strong>of</strong> the pro<strong>of</strong> <strong>of</strong> this theorem follows<br />

from Theorem 8.6.19.<br />

(c) Provide the missing component to complete the pro<strong>of</strong> <strong>of</strong> your theorem.<br />

EXERCISE 9.15.5 Suppose R and S are rings and φ : R → S is an epimorphism.<br />

Then S ∼ = R/ Ker(φ).<br />

Let’s return to the morphism in Exercise 9.14.9 defined by φ(n) = ne. If a ring<br />

has nonzero characteristic, then there is a smallest positive integer n for which<br />

ne = 0. By Exercise 9.14.19, Ker(φ) is an ideal in the integers, which is a PID.<br />

Thus Ker(φ) = (k) for some integer k. Since φ(k) = 0, and since n is the smallest<br />

positive integer for which ne = 0, it must be that k ≥ n. But since n ∈ Ker(φ), it<br />

must be that n is a multiple <strong>of</strong> k, so that k ≤ n. Thus k = n. By Exercise 9.15.5, the<br />

range <strong>of</strong> φ is isomorphic to Z/(n). That is, R contains a subring isomorphic to Zn.<br />

If we think <strong>of</strong> a ring with unity e as an abelian additive group and let S be the<br />

subgroup generated by e, the identity element <strong>of</strong> the other operation, then the<br />

form <strong>of</strong> S can be determined by looking at the additive form <strong>of</strong> Eq. (8.22).<br />

S ={ne : n ∈ Z} (9.71)<br />

which is precisely the range <strong>of</strong> φ. Thus as far as addition is concerned, S is the<br />

smallest additive subgroup <strong>of</strong> R that contains e. If we could show that S is closed<br />

under multiplication, we would have that S is the smallest subring <strong>of</strong> R that contains<br />

e. But closure is immediate by Exercise 9.3.20. In bits and pieces, we have<br />

proved the following.<br />

Theorem 9.15.6 Suppose R is a ring with unity. If R has characteristic zero,<br />

then it contains a subring isomorphic to the integers. If R has positive characteristic<br />

n, then it contains a subring isomorphic to Zn. In either case, such is the smallest<br />

subring <strong>of</strong> R that contains e.

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