Proof exercise 2 an injective homomorphism f r r is

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Proof. Exercise. 2 An injective homomorphism f : R R 0 is called an embedding of R in R 0 . In this case, f ( R ) is a subring of R 0 and R = f ( R ), and we say that “ R is embedded in R 0 ,” or as an abuse of terminology, one might simply say that “ R is a subring of R 0 .” Theorems 4.18, 4.19, and 4.20 also have natural analogs: Theorem 5.15 If I is an ideal R , then the map f : R R/I given by f ( a ) = a + I is a surjective homomorphism whose kernel is I . This is sometimes called the “natural” map from R to R/I . Proof. Exercise. 2 Theorem 5.16 Let f be a homomorphism from R into R 0 . Then the map ¯ f : R/ ker( f ) f ( R ) that sends the coset a + ker( f ) for a R to f ( a ) is unambiguously defined and is an isomorphism of R/ ker( f ) with f ( R ) . Proof. Exercise. 2 Theorem 5.17 Let f be a homomorphism from R into R 0 . The ideals of R containing ker( f ) are in one-to-one correspondence with the ideals of f ( R ) , where the the ideal I in R containing ker( f ) corresponds to the ideal f ( I ) in f ( R ) . Proof. Exercise. 2 42

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Example 5.16 For n > 1, the natural map f from Z to Z n sends a Z to the residue class [ a mod n ]. This is a surjective map with kernel n Z . Consider the multiplicative group of units Z * = 1 } and the restriction f * of f to Z * . This is a homomorphism from Z * into Z * n with kernel ker( f * ) = (1 + n Z ) ∩ {± 1 } . Thus, if n = 2, ker( f * ) = 1 } , and otherwise, ker( f * ) = { 1 } . 2 Example 5.17 We may restate the Chinese Remainder Theorem (see Theorem 2.6) in more al- gebraic terms. Let n 1 , . . . , n k be integers, all greater than 1, such that gcd( n i , n j ) = 1 for all 1 i < j k . Consider the homomorphism from the ring Z to the ring Z n 1 × · · · × Z n k that sends x Z to ([ x mod n 1 ] , . . . , [ x mod n k ]). In our new language, Theorem 2.6 says that this homomorphism is surjective and the kernel is n Z , where n = Q k i =1 n i . Therefore, the map that sends [ x mod n ] Z n to ([ x mod n 1 ] , . . . , [ x mod n k ]) is an isomorphism of the ring Z n with the ring Z n 1 ×· · ·× Z n k . The restriction of this map to Z * n yields an isomorphism of Z * n with Z * n 1 ×· · ·× Z * n k . 2 Example 5.18 Let n 1 , n 2 be positive integers with n 1 > 1 and n 1 | n 2 . Then the map f : Z n 2 Z n 1 that sends [ a mod n 2 ] to [ a mod n 1 ] is a surjective homomorphism, [ a mod n 2 ] ker( f ) if and only if n 1 | a , i.e., ker( f ) = n 1 Z n 2 . 2 Example 5.19 Fix x R . The map that sends a R [ T ] to a ( x ) is a homomorphism from R [ T ] onto R . The kernel is the ideal generated by ( T - x ). Thus, R [ T ] / ( T - x ) = R . 2 Example 5.20 Let us continue with Example 5.15. The map f : R S that sends r R to [ r mod m ] S is an embedding of R in S . 2 Example 5.21 For any ring R , consider the map f : Z R that sends m Z to m · 1 R in R . This is clearly a homomorphism of rings. If ker( f ) = { 0 } , then the ring Z is embedded in R , and R has characteristic zero. If ker( f ) = n Z for n > 0, then the ring Z n is embedded in R , and R has characteristic n . 2 43
Chapter 6 Polynomials over Fields Throughout this chapter, K denotes a field, and D denotes the ring K [ T ] of polynomials over K . Like the ring Z , D is an integral domain, and as we shall see, because of the division with remainder property for polynomials, D has many other properties in common with Z as well.

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• Spring '13
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