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Math 412. the First Isomorphism Theorem

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Math 412. the First Isomorphism Theorem (c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. The First Isomorphism Theorem. φ NOETHER’S FIRST ISOMORPHISM THEOREM: Let R −! S be a surjective homomor- phism of rings. Let I be the kernel of φ. Then R=I is isomorphic to S. More precisely, there is a well-defined ring isomorphism R=I ! S given by r + I 7! φ(r). A. WARM-UP: (1) Prove that the kernel of any ring homomorphism φ : R ! S is an ideal of the source ring. (2) Prove that the image of any ring homomorphism φ : R ! S is a subring of the target ring. (1) This is proved on page 155 (Theorem 6.10). Actually, here is a better proof. We need to show that ker φ is closed under addition and under both left and right multi- plication by arbitrary element of R. Say x; y 2 ker φ. To check that x + y 2 ker φ, apply φ(x + y) = φ(x) + φ(y) = 0 + 0 = 0. So x + y 2 ker φ. Take r 2 R arbi- trary. To check that rx is in ker φ, apply φ: φ(rx) = φ(r)φ(x) = φ(r) · 0 = 0, so rx 2 ker φ. Likewise, applying φ to xr, we haveφ(xr) = φ(x)φ(r) = 0 · φ(r) = 0, so xr 2 ker φ. QED (2) This is proved on page 77 of the text (Corollary 3.11). We do have one extra thing to check, since our definition of rings (and subrings) include the multiplicative identity. We check that 1S 2 im(φ), which is easy, since φ(1R) = 1S by definition of ring homomorphism, so 1S is in the image of φ. B: Consider the canonical homomorphism: Z ! Zn a 7! [a]n (1) Make sure everyone in your group can explain why this map is a surjective ring homo- morphism. (2) Compute the kernel. (3) Find a generator for the kernel. (4) Verify the first isomorphism theorem for this ring homomorphism. (1) We’ve proved this multiple times but no shame if it is not clear: Please come talk to me if you are unsure. (2) The kernel is the ideal of multiples of n. (3) A generator is n (another generator is −n). (4) The first isomorphism theorem says that the quotient ring Z=(n) is isomorphic to Zn. This is indeed true: you proved it on the last worksheet in the first problem. Even more than isomorphic: these are literally the same set of sets (congruence classses mod n). C: Fix any real number a. Consider the evaluation map η : R[x] ! R f 7! f(a) (1) Make sure you understand why the evaluation map is a surjective ring homomor- phism. 1 (2) Prove that the kernel of η is the ideal I of R[x] generated by (x − a). (3) Use the first isomorphism theorem to prove that R[x]=(x − a) is isomorphic to R. ∼ (4) Give a direct proof that R[x]=(x − a) = R by thinking about the congruence classes [f](x−a). Why is there a bijection with R that preserves the ring structure? (1) Please come talk to me if you are unsure why this is a ring homomorphism. To prove it is surjective: take arbitrary λ 2 R (the target). Let f(x) 2 R[x] (the source) be the constant polynomial f(x) = λ. Then the evaluation map sends f to λ. So it is surjective. (2) The kernel is ff 2 R[x] j f(a) = 0g. By the Remainder theorem, f(a) = 0 if and only if (x − a) is a factor of f. This means the kernel is f(x − a)g(x) j g 2 R[x]g. So (x − a) is a generator. (3) Apply the first isomorphism theorem to the evaluation map R[x] ! R f 7! f(a): The kernel is (x − a) so the quotient ring R[x]=(x − a) =∼ R. (4) Each class in R[x]=(x − a) is uniquely represented by the common remainder of the elements in it when divided by (x − a) (from worksheet last time). These remainders have degree less than 1 (the degree of x − a). So they are constant polynomials. So there is a So when we add/multiply classes in R[x]=(x − a), we can represent them all by [λ]x−a where λ 2 R. The way we add/multiply is “the same” as the way we add/multiply in R. That is: [λ]x−a · [µ]x−a = [λµ]x−a and similarly for addition. Clearly this is a bijection: each λ 2 R lies in exactly one class [λ]x−a and each class [g]x−a contains exactly one constant polynomial (which we know from the remainder theorem to be g(a).) When we identify the class of g with the remainder after dividing by (x − a); we get f(a) by the Remainder theorem. This is the evaluation map! p D: Let i be the complex number −1. Consider the ring homomorphism φ : R[x] ! C f 7! f(i) (1) Prove that φ is surjective. (2) Prove that x2 + 1 2 ker φ. (3) Prove that the kernel contains no (non-zero) polynomial of degree less than two. (4) Prove that x2 + 1 generates ker φ. [Hint: If f(x) is in the kernel, use the division algorithm to divide f by x2 + 1 and see what happens under φ.] (5) Use the First Isomorphism Theorem to explain how to think about the complex numbers as a quotient of the polynomial ring R[x]: (1) Take arbitrary a+bi in the target C. The polynomial a+bx in the source satisfies φ(a+bx) = a+bi, so φ is surjective. (2) φ(x2 + 1) = i2 + 1 = 0. So x2 + 1 2 ker φ. (3) Take arbitrary a + bx of degree one or less. Apply φ to get φ(a + bx) = a + bi which is not the zero complex number unless a = b = 0. So no non-zero element of degree one or less is in the kernel. (4) We need to prove ker φ = (x2 + 1): First we show (x2 + 1) ⊂ ker φ. Take arbitrary g(x)(x2 + 1) 2 (x2 + 1): SInce the kernel is an ideal and x2 + 1 2 ker φ, every multiple of x2 + 1 2 ker φ. So (x2 + 1) ⊂ ker φ. For the other direction: Take f 2 ker φ. Use the division algorithm to write f = (x2 + 1)q(x) + r(x) where r = 0 or degree r < 2. We need to show r = 0. Apply φ: φ(f) = φ(x2 + 1)φ(q) + φ(r): SInce f and x2 + 1 are in the kernel, we see that 0 = 0φ(q) + φ(r). So φ(r) = 0. But the kernel has no element of degree 1 or less! So r = 0. This means f 2 (x2 + 1). So ker φ ⊂ (x2 + 1). (5) The map φ is a surjective ring homomorphism and the kernel is (x2 + 1): So the first isomorphism theorem says that C =∼ R[x]=(x2 + 1): E: Let n and m be integers. 1One direction is easy. For the other, say g 2 ker η, and use the division algorithm to divide g by (x − a). Then apply η. (1) There is a unique ring homomorphism φ : Z ! Zn × Zm. Find it. Why is it unique? (2) Prove that the kernel is generated by the least common multiple of m and n. In par- ticular, if m and n are relatively prime, explain why the kernel consists of multiples of mn. (3) Let d be the least common multiple of m and n. Use the first isomorphism theorem to ∼ prove that there is a ring isomorphism Zd = im(φ). (4) Explain why if m and n are relatively prime your ring homomorphism φ is surjective.2 (5) Use the first isomorphism theorem to prove that if m and n are relatively prime, then ∼ Zmn = Zn × Zm. [You gave a different proof on Problem Set 5.] (1) Define φ : Z ! Zn × Zm by φ(x) = ([x]n; [x]m). It is easy to check that this is an homomorphism (do it!). Note that because 1 7! ([1]n; [1]m), the entire mapping is completely determined! Every element of Z is either 1 + 1 + ··· + 1 (x times say, to get x) or the additive inverse of this, and ring homomorphisms respect addition, we know x = 1 + 1 + ··· + 1 MUST go to ([1]n; [1]m) + ([1]n; [1]m) + ··· + ([1]n; [1]m) = ([x]n; [x]m) (and similarly for −x). This argument shows that there is ONLY ONE homomorphism from Z to any ring, in fact. (2) An integer a is mapped to ([0]n; [0]m) if and only if ([0]n; [0]m) = ([a]n; [a]m), which is to say: a is a multiple of both m and n. Every multiple of both m and n is a multiple of the least common multiple, so the least common multiple generates the kernel. If m and n are relatively prime, then the lcm is mn, as follows from the fundamental theorem of arithmetic. (3) Since the image of φ is a ring, the map Z ! im(φ) sending x 7! ([x]n; [x]m) is a surjective ring ∼ homomorphism. Its kernel is generated by the lcm d. So Zd = im(φ). (4) To show φ is surjective, take any ([a]; [b]) 2 Zn × Zm.
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