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(c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Examples and Types of Rings and their Homomorphisms DEFINITION:A domain is a commutative ring R in which 0R 6= 1R, and which has the property that whenever ab = 0 for a; b 2 R, then either a = 0 or b = 0. DEFINITION:A field is a commutative ring R in which 0R 6= 1R in which every non-zero element has a multiplicative inverse. DEFINITION:A subring1 S of a ring R is a subset of R satisfying the following (1) 1R 2 S; (2) S is closed under multiplication; (3) S is closed under addition and taking additive inverses. Equivalently, a subring S of a ring R as a subset which is a also a ring with the same +; ×; 0 and 1. “DEFINITION:” Fix a commutative ring R. The polynomial ring over R is the set n R[x] = fa0 + a1x + ··· + anx j ai 2 R; n 2 Ng; with operations + and × extended from those on the coefficients in R in the natural way. THEOREM 4.3: Fix a ring R. Then the polynomial R[x] is a domain if and only if R is a domain. THEOREM 4.5: For any domain R, the units in R[x] are the units in the subring R of constant polynomials. In particular, if F is a field, then the units in F[x] are the non-zero constant polynomials. A. WARM-UP: For each inclusion S ⊂ R, set whether or not S subring of R. (1) Z ⊂ Q; Z ⊂ R; R ⊂ C. (2) N ⊂ Z, (3) The set of even integers S = f2n j n 2 Zg ⊂ Z. f(x) (4) R[x] ⊂ R(x) := f g(x) jf(x); g(x) 2 R[x]; g 6= 0g (the set of rational functions). a 0 (5) The set of diagonal matrices D := f j a; b; 2 g ⊂ M ( ). 0 b R 2 R (6) The set of integer matrices M2(Z) ⊂ M2(R). a b (7) The set of determinant 1 matrices GL ( ) = f j ad − bc = 1; 2 g ⊂ M ( ). 2 R c d R 2 R (8) The set of constant polynomials R ⊂ R[x]. (9) The set of polynomials with integer coefficients Z[x] ⊂ R[x]. (10) Z ⊂ Z[i] (11) The imaginary integers Zi = fni j n 2 Zg ⊂ Z[i]. (1) All yes. (2) No, no additive inverses. (3) No. missing multiplicative identity. (4) Yes. (5) yes. (6) yes. (7) No, no zero. (8) Yes. (9) YEs. (10) Yes. (11) No, no 1. B. FIND AN EXAMPLE OF: 2 (1) A non-commutative ring, with a commutative subring. (2) An infinite ring with a finite subring. (3) A field which has a subring which is not a field. For (1) use (5) above. For (2), use Z2 ⊂ Z2[x]. For (3), use Z ⊂ Q. C. BASIC PROOFS. (1) Prove that every field is a domain. (2) Prove that a subring of a field is a domain. Is the converse true? (3) Let S be a subset of a ring R. Explain why S is a subring if and only if the inclusion map S,! R sending s 7! s is a ring homomorphism. For this, you don’t have to write a lot, but think carefully about the meaning of the symbols you are using in different contexts. (1) Let F be a field. Assume a; b 2 F satisfy ab = 0 but a 6= 0. Multiplication on the left by a−1 gives a−1(ab) = (a−1a)b = 1 · b = b = 0: QED. (2) It is clear that a subring of a domain is a domain: assume a; b 2 S ⊂ R, where R is a domain, and ab = 0 in S. This also holds in the bigger ring R, so a = 0R or b = 0R, since R is a a domain. Since 0R = 0S, it follows that S is a domain too. (3) Call the inclusion map φ. First, assume S ⊂ R is a subring. We need to prove φ is a ring homo- morphism. Since the 1 in R is the 1 of S, we know φ(1S) = 1R. Also since φ is the inclusion map, φ(s1 + s2) = s1 + s2 = φ(s1) + φ(s2): Ditto for muiltiplication. So φ is a ring homomorphism. Conversely, assume that φ : S,! R is a ring homomorphism. In particular S is a ring. Then 1S = φ(1S) = 1R, so S and R have the same identity. Also φ(s1 +S s2) = φ(s1) +R φ(s2) = s1 +R s2. This is in S, since it is in the image of s1 +S s2 under φ. So S is closed under the multiplication of R. Similarly, S is closed under the multiplication of R. Finally, for all s 2, we have s +R −s = 0S using the addition in S. Applying φ, we have φ(s +R −s) = φ(0S), which means that φ(s) +R Sφ(−s) = 0R. D. POLYNOMIAL RING PRACTICE. Answer the following questions, using Theorem 4.5 and 4.5 above where appropriate. (1) What is the zero and one in Z3[x]? Is Z3[x] commutative? Is it a domain? (2) Use a theorem to investigate if Z72[x] a domain? If not, find a pair of non-zero elements in Z72[x] whose product is zero. 2 3 3 (3) In Z8: for f = (1+3x) and g = (2x +4x ), compute and simplify f +4g and (3x) +g. Here, we abuse notation by representing congruence classes by any integer representative. (4) How many polynomials of degree less than 3 in the ring Z2[x]? (5) How many units in Z[x]? (6) Suppose that f 2 Q[x] has degree 5. Find the degrees of the following polynomials: f − x, f 2, f + 4x51; f − 2x5, (x2 + 1)f 3. 7 16 15 (7) Find the leading coefficient of h where h = 5x + 3x + 1 2 Z7[x]. 2 (8) Does x + 1 have a multiplicative inverse in Z2[x]? (9) Is there any field F such that F[x] is a field? (10) In Z8[x], compute (1 + 4x)(1 − 4x). Is the hypothesis that R is a domain necessary in Theorem 4.5? (1) The zero and one are the constant polynomials [0]3 and [1]3. Yes, F3[x] is commutative and a domain by Theorem 4.3. 3 2 (2) Theorem 4.3 implies Z72[x] is not a domain. Note: [8]72x · ([9]x ) = 0. 3 3 2 (3) In Z8: f + 4g = 1 + 3x and (3x) + g = 7x + 2x . (4) There are 7 polynomials of degree less than 3 in the ring Z2[x], if we go with the convention that the 0 polynomial has undefined degree. There are 8 if we go with the convention that the 0 polynomial has degree −∞. Only 1 is a unit. (5) There are two units in Z[x]; ±1, by Theorem 4.5. (6) deg(f − x) = 5, deg(f 2) = 10, deg(f + 4x51) = 51, deg(f − 2x5) is 5 if the leading coefficient of f is not 2, less otherwise. deg((x2 + 1)f 3) = 17. 16 15 7 7 (7) The leading coefficient of (5x + 3x + 1) 2 Z7[x] is 5 = 5. (8) Nope. Thm 4.5. (9) Nope. Thm 4.5 says x is never a unit. (10) In Z8[x], (1 + 4x)(1 − 4x) = 1. So the hypothesis that R is a domain necessary in Theorem 4.5. E. PROOF OF THEOREM 4.5 Let R be a domain. Consider R as the subring of constant polynomials in R[x]. (1) Show that any unit in R is a unit in R[x]. (2) Explain why, for any two polynomials f; g 2 R[x], deg(fg) = deg f + deg g. Is this true even when R is not a domain? (3) Prove that if f 2 R[x] is a unit, then f is a constant polynomial. (4) Prove Theorem 4.5 (5) Find a formula for the number of units in Zp[x] where p is prime. This is all in the book, in Section 4.1. For (1), just show directly that if u is a unit in R, with inverse v, then uv = 1 interpreted in R or in R[x]. For (2) see Theorem 4.2. It is false when R is not a domain: 2x · (3x + 1) = 2x in Z6[x]. For (3) and (4), read the proof of Theorem 4.5 in the book. For (5), the theorem × × says that (Zp[x]) = Zp . Since Zp is a field, all elements (except 0) are units, so there are p − 1. F. HOMOMORPHISMS. Let F be an arbitrary field. Fix any a 2 F. (1) Prove that the map F[x] ! F sending f(x) 7! f(a) is surjective homomorphism. Write the kernel in set-builder notation. (2) Find an injective homomorphism F ! M2(F). Also find an injective map F ! M2(F) which is not a ring homomorphism. (3) Is the map M2(F) ! F × F sending a matrix A to its first row a ring homomorphism? G. THE DIVISION ALGORITHM FOR POLYNOMIALS. In high school, you learned to divide polynomials to find a quotient and remainder. This works for polynomials over any field,. Let’s practice. (1) Let f = x3 + 4x2 + x + 1 in R[x]. Find q and r so that f = qx2 + r, where deg r < 2.
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