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R → R Be a Ring Homomorphism Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 3, 2006 1 TALK SLOWLY AND WRITE NEATLY!! 0.1 Rings Today we are going to introduce a new structure which you haven't seen before. They are called Rings and they are a generalization of the integers. 0.1.1 De¯nitions De¯nition of Ring De¯nition 0.1.1.1. We say hR; +; £; 0; 1i is a Ring if ² R is a set with 0; 1 2 R ² +; £ : R £ R ! R ² R+ = hR; +; 0i is an abelian group ² (8x; y; z 2 R)(x £ y) £ z = x £ (y £ z) 2 ² (8x 2 R)x £ 1 = 1 £ x (Distributive Laws) (8x; y; z 2 R)x £ (y + z) = x £ y + x £ z (8x; y; z 2 R)(y + z) £ x = y £ x + z £ x We say a ring is Commutative if (8a; b 2 R) De¯nition of Subring De¯nition 0.1.1.2. Let hR; +; £; 0; 1i be a ring. We say that S ⊆ R is a Subring of hR:+; £; 0; 1i if 0; 1 2 S (8x; y 2 S)x + y 2 X ^ (¡x) 2 S ^ x £ y 2 S 0.1.2 Examples Integers Integers: 3 One of the most important examples of a commutative ring is the ring of integers Z. Some properties of the ring of integers which are inter- esting are ² Z is commutative. ² Z has no subrings. This is because if S ⊆ Z is a subring then it contains 0; 1 and hence contains 1 + 1 + ¢ ¢ ¢ + 1 n times for all n. And similarly contains ¡(1 + ¢ ¢ ¢ + 1) and hence contains all the integers. Gaussian Integers Gaussian Integers: 4 De¯nition 0.1.2.1. We say that Z[i] = fa + bi : a; b 2 Zg is the ring of Gaussian Integers The Gaussian integers are the points of a square lat- tice in the complex plane. DRAW THE COMPLEX PLANE WITH THE POINTS OF THE GAUSSIAN IN- TEGERS. Subring of C generated by ®: Similar to the ring of Gaussian integers, given any com- plex number ® 2 C we can form i Z[®] = f§i·nai® :(8i)ai 2 Z and n 2 f0; 1; 2;::: gg Notice that Z[®] is the smallest subring of C containing ®. We call Z[®] the subring generated by ®. ASK IF THEY WANT ME TO PROVE THAT Z[®] IS THE SMALLEST SUBRING CONTAINING ®. 5 Algebraic/Transcendental Number De¯nition 0.1.2.2. We say that ® 2 C is algebraic if 1 2 n (9a0; a1; : : : ; an 2 Z)a0 + a1® + a2® + ¢ ¢ ¢ + an® = 0 We say that ® is transcendental if it is not algebraic. p pp Some examples of algebraic numbers are i; 2; 2 + 17 while some examples of transcendental numbers are e; ¼. If ® is transcendental then there is a bijection p(®) à p(x) From the ring generated by ® to the ring of polynomials with coe±cients in Z. Polynomial Rings Ring of Polynomials over a Ring: In general we have that if R is a ring then i R[x] = f§i·naix : ai 2 Rg 6 is also a ring. Zero Ring Zero Ring: De¯nition 0.1.2.3. We de¯ne hf0g; +; ¤; 0; 0i to be the Zero Ring We then have that Theorem 0.1.2.4. If R is a ring such that 0 = 1 then R is the zero ring. Proof. First note that 0a = 0 for any element of a ring. This is because 0a + 0a = (0 + 0)a = 0a so subtracting 0a from both sides we see 0a = 0. Now assume 0 = 1 in a ring R. Then for all a 2 R, a = 1a = 0a = 0. Fields Fields: 7 Every ¯eld is also a ring. In fact ¯elds are special types of rings because every element other than 0 has a multiplica- tive inverse. There is a special name for such elements. De¯nition of Unit De¯nition 0.1.2.5. Let hR; £; +; 0; 1i be a ring. We say that u 2 R is a unit of R if (9u¡1 2 R)u £ u¡1 = u¡1 £ u = 1 So a ¯eld is a commutative ring where every element other than 0 is a unit. Matrixes Real Matrixes: Our last example of a ring are the real n £ n matrixes. This is an important example of a ring because it is an example of a non-commutative ring. 0.2 Homomorphisms and Ideals 0.2.1 De¯nitions Ring Homomorphism 8 De¯nition 0.2.1.1. Let hR; £; +; 0; 1i; hR0; £0; +0; 00; 10i be rings. A Homomorphism from R to R0 is a map Á : R ! R0 such that Á(0) = 00;Á(1) = 10;Á(a+b) = Á(a)+0Á(b);Á(a£b) = Á(a)£0Á(b) for all a; b 2 R. We say that Á is an isomorphism if it is a bijective ho- momorphism. If there is an isomorphism between R; R0 then we say R and R0 are isomorphic 0.2.2 Theorems The most important ring homomorphisms are those which are obtained by evaluating a polynomial at a value. For example, if a 2 R then we have Á : p(x) à p(a) is a homomorphism Á : R[x] ! R. 9 This leads to the following theorem. Substitution Principle Theorem 0.2.2.1 (Substitution Principle). Let ' : R ! R0 be a ring homomorphism. (a) Given an element ® 2 R0 there is a unique ho- momorphism © : R[x] ! R0 which agrees with the map ' on constant polynomials and sends x à ® 0 (b) More generally, given elements ®1; : : : ; ®n 2 R there is a unique homomorphism © : R[x1; : : : ; xn] ! R0 from the polynomial ring in n over R to R0 which agrees with ' on constant polynomials and which sends xi à ®i Proof. First note that the proof of (b) is obtained by sim- ply repeatedly applying (a). 10 Note that if © exists then we must for each polynomial that i i ©(§rix ) = §'(ri)® because we have determined where © send coe±cients of the polynomial as well as where it sends x (and it is a ring homomorphism. Hence, if © exists it must be unique. To prove its existence lets take the above formulas as the de¯nition for © and show that it is in fact a homo- morphism from R[x] to R0. ² Note that ©(1) = '(1) = 1. ² It is obviously true that © commutes with addition i j ² Let f = §iaix ; g = §jbjx . Then i i+j i j ©(fg) = ©(§i;jaibjx +j) = §i;j'(ai)'(bj)® = (§i'(ai)® )(§j'(bj)® = ©(f)©(g) 11 Corollary 0.2.2.2. Let x = (x1; : : : ; xn); y = (y1; : : : ; ym) denote sets of variables. Then there is a unique mor- phism ' : R[x; y] ! R[x][y] which is the identity on R [ x [ y. Proof. Note that R is a subring of R[x] and R[x] is a subring of R[x][y] so R is a subring of R[x][y]. Let ' : R ! R[x][y] be the inclusion map. The substitu- tion principle tells us that there is a unique extension of ' to a map from © : R[x; y] ! R[x][y] sending x to x and y to y while agreeing with ' on R. However, we also know that there is an inclusion map à : R[x] ! R[x; y]. And so by the substitution principle there must be a unique extension ª : R[x][y] ! R[x; y] sending y to y and agreeing with à on R[x]. 12 But then ª© and ©ª are both the identity on R and on x; y so they must be the identity on their respective domains. And hence inverse isomorphisms. We also have a relationship between polynomials as ele- ments of R[x] and polynomials as functions via the above substitution maps. This can be seen by the following the- orem. Theorem 0.2.2.3. Let R denote the ring of contin- uous real-valued functions on Rn. The map ' : R[x1; ¢ ¢ ¢ ; xn] !R sending a polynomial to its associated polynomial func- tion is an injective homomorphism. Proof. Immediate. Another set of important ring homomorphisms comes from the following Integers are an Initial Object 13 Theorem 0.2.2.4. For every ring R there is exactly one homomorphism ' : Z ! R It is the map which takes '(n) = 1R + 1R + ¢ ¢ ¢ + 1R n times if n > 0 and '(¡n) = ¡'(n). Proof. First we will show uniqueness. Suppose there are two homomorphisms '; ÃZ ! R. Then we must have '(1) = 1R = Ã(1). Now lets assume ' and à agree on all numbers between x, ¡n · x · n. Then '(n + 1) = '(n) + '(1) = Ã(n) + Ã(1) = Ã(n + 1). And so by induction ' = psi on Z. Now lets consider the map de¯ned by ² '(1) = 1R ² '(n + 1) = '(n) + 1 ² '(¡n) = ¡'(n) 14 This is the only possible homomorphism from Z ! R. So all that is left is to show that it is an actual homomor- phism. Addition: First notice that '(1 + 1) = '(1) + '(1). Now assume that if m + n = r (and both are positive) then '(m + n) = '(m) + '(n). Now let s + t = r + 1. Then we have '(s+t) = '(r+1) = '(r)+'(1) = '(s+(t¡1))+'(1) = '(s)+'(t¡1)+'(1) = '(s)+'(t) And so by induction we have ' commutes with addition. Multiplication: First notice that '(1 ¤ r) = '(1) ¤ '(r) = '(r). Now assume that '(m ¤ n) = '(m) ¤ '(n). We then have '((m+1)¤n) = '(m¤n+m) = '(m)¤'(n)+'(m) = ('(m)+'(1))¤'(n) = '(m+1)¤'(n) 15 And so by induction (and the previous result concerning addition) we have ' commutes with multiplication and hence is a homomorphism.
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