Math 412. the First Isomorphism Theorem

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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. The First Isomorphism Theorem.

φ NOETHER’S FIRST ISOMORPHISM THEOREM: Let R −→ S be a surjective homomorphism 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 multiplication by arbitrary element of R. Say x, y ∈ ker φ. To check that x + y ∈ ker φ, apply φ(x + y) = φ(x) + φ(y) = 0 + 0 = 0. So x + y ∈ ker φ. Take r ∈ R arbitrary. To check that rx is in ker φ, apply φ: φ(rx) = φ(r)φ(x) = φ(r) · 0 = 0, so rx ∈ ker φ. Likewise, applying φ to xr, we haveφ(xr) = φ(x)φ(r) = 0 · φ(r) = 0, so xr ∈ 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 ∈ 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 homomorphism. (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 homomorphism. 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 ∈ R (the target). Let f(x) ∈ 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 {f ∈ R[x] | f(a) = 0}. By the Remainder theorem, f(a) = 0 if and only if (x − a) is a factor of f. This means the kernel is {(x − a)g(x) | g ∈ R[x]}. 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 ∈ 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 ∈ 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!

√ 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 ∈ 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 ∈ 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) ∈ (x2 + 1). SInce the kernel is an ideal and x2 + 1 ∈ ker φ, every multiple of x2 + 1 ∈ ker φ. So (x2 + 1) ⊂ ker φ. For the other direction: Take f ∈ 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 ∈ (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 ∈ 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]) ∈ Zn × Zm. We need to ﬁnd x ∈ Z such that [x]n = [a]n and [x]m = [b]m. That is, we need x ∈ Z such that x ≡ a mod n and y ≡ b mod m. The Chinese Remainder Theorem exactly says that when (m, n) = 1, such an x always exists! So the map is surjective. (5) This is a special case of (3), noting that the lcm is mn and the image is the full target because the map is surjective when m and n are coprime.

F:PROOFOFTHE FIRST ISOMORPHISM THEOREM. Fix a surjective ring homomorphism φ : R → S. Let I be its kernel. Deﬁne φ : R/I → S by φ(r + I) = φ(r). (1) Show that φ is a well-deﬁned map. (2) Show that φ is a surjective ring homomorphism. (3) Show that φ is injective. (4) Prove the First Isomorphism Theorem.

This is on page 157 in the text. The proof of Theorem 6.13.

G: Let f ∈ F[x] be a polynomial of degree d. Consider the quotient ring R = F[x]/(f). (1) Suppose f factors as f = gh where g and h have strictly smaller degree than f. Prove that the congruence class g + (f) is a zero divisor in R. (2) State and prove a necessary and sufﬁcient condition for R = F[x]/(f) to be a domain. (3) Prove that if f is irreducible and g is arbitrary, then the greatest common divisor of f and g is 1 unless f|g. (4) Prove that if the greatest common divisor of f and g is 1, then the class g + (f) is a unit in R = F[x]/(f). Clearly state the relevant theorems you use. (5) Prove that R = F[x]/(f) is a ﬁeld if and only if f is an irreducible polynomial.

2Recall you proved the Chinese Remainder Theorem on Problem Set 5: Fix m and n relatively prime integers. For any a, b ∈ Z, there is always a solution to the pair of congruences x ≡ a mod n and x ≡ b mod m. (1) gh = f means [gh] = [g][h] = [f] = [0]. Note that since g, h have degree less than f, neither one is in (f), so the classes [g] = g + (f) and [h] = h + (f) are non-zero. So [g] = g + (f) is a zero divisor. (2) R = F[x]/(f) is a domain if and only if f is irreducible. Proof: Say f is NOT irreducible. Then f = gh, where g, h are not constant polynsomials, so by (1), [g][h] = 0 and R is not a domain. For the converse, say f is irreducible. Take classes [g] and [h] in R = F[x]/(f) whose product is zero. This means [g][h] = [gh] = 0. So gh ∈ (f). So f|(gh). But f is IRREDUCIBLE, so it must divide g or h. So [g] or [h] is zero. (3) See Theorem 5.10 on page 135 in the text for (3), (4) and (5). This is an exact analog of the corre- sponding fact for Z: namely Zn is a ﬁeld if and only if n is prime.

H: Let S ⊂ Q be the subset of rational numbers with odd denominators (when expressed in lowest terms). (1) Prove that S is a subring of Q. (2) Let I ⊂ S be the subset of rational numbers with odd denominator and even numerator (when expressed in lowest terms). Prove that I is an ideal of S. (3) Show that the quotient ring S/I is isomorphic to Z2. [Hint: The elegant proof uses the ﬁrst isomorphism theorem.]

a b (1) Take 2n+1 , 2m+1 ∈ S. We need to show that a b a b a(2m+1)+b(2n+1) (a) 2n+1 + 2m+1 ∈ S. This is true: 2n+1 + 2m+1 = (2n+1)(2m+1) ∈ S. Note that the denominator is odd. a b a b ab (b) 2n+1 · 2m+1 ∈ S. This is true: 2n+1 · 2m+1 = (2n+1)(2m+1) ∈ S. a (c) − 2n+1 ∈ S, obvious. 1 (d) 1 = 1 ∈ S. QED. (2) We need to show 2a 2b 2a 2b 2a 2b (a) If 2n+1 , 2m+1 ∈ I, then also 2n+1 + 2m+1 ∈ I. This is true: 2n+1 + 2m+1 = 2[a(2m+1)+b(2n+1)] (2n+1)(2m+1) . 2a b 2a b 2a b (b) For 2n+1 ∈ I and 2m+1 ∈ S, we have 2n+1 · 2m+1 ∈ I. Also clear: 2n+1 · 2m+1 = 2ab (2n+1)(2m+1) ∈ I. So I is an ideal. ∼ (3) S/I = Z2. The elegant argument use the first isomorphism theorem: Define a homomorphism S → Z2 which takes the even-numerator-fractions to [0] and the odd-numerator-fractions to [1]. It is easy to see this is a homomorphism, surjective with kernel I. So S/I = Z2. Alternatively, here is another more hands on argument. First observe that there are two congruence classes, 0 + I, which is just I, the fractions with even numerator and odd denominator (in lowest terms), and 1 + I, which consists of the fractions with odd numerator and odd denominator (in lowest terms). So the quotient ring S/I has two elements, 1 + I and 0 + I. This is clearly isomorphic to Z2. We can define a map S → Z2 which takes the class of even-numerator-fractions to [0] and the class of odd-numerator- fractions to [1]. To check that this map preserves addition and multiplication you can write out the tables and observe that they are the same after this nenaming.

I:PRIME IDEALS An ideal I in a commutative ring R is prime if whenever rs ∈ I, it follows that either r ∈ I or s ∈ I. (1) Prove that I is prime if and only if R/I is a domain. (2) Find all prime ideals in Z. (3) Find all prime ideals in R[x]. (1) is Theorem 6.14 on page 163. See the proof there. Prime ideals in Z are the zero-ideal and the ideals (p) of multiples of a prime number. Prime ideals in F[x] are the zero-ideal and the ideals (f(x)) of multiples of an irreducible polynomial.