Are Maximal Ideals? Solution. No Principal

Math 403A Assignment 6. Due March 1, 2013.

1.(8.1) Which principal ideals in Z[x] are maximal ideals?

Solution. No principal ideal (f(x)) is maximal. If f(x) is an integer n = 1, then (n, x) is a bigger ideal that is not the whole ring. If f(x) has positive degree, then ± take any prime number p that does not divide the leading coeﬃcient of f(x). (p, f(x)) is a bigger ideal that is not the whole ring, since Z[x]/(p, f(x)) = Z/pZ[x]/(f(x)) is not the zero ring.

2.(8.2) Determine the maximal ideals in the following rings. (a). R R. × Solution. Since (a, b) is a unit in this ring if a and b are nonzero, a maximal ideal will contain no pair (a, 0), (0, b) with a and b nonzero. Hence, the maximal ideals are R 0 and 0 R. ×{ } { } × (b). R[x]/(x2).

Solution. Every ideal in R[x] is principal, i.e., generated by one element f(x). The ideals in R[x] that contain (x2) are (x2) and (x), since if x2 R[x]f(x), then x2 = f(x)g(x), which 2 ∈ 2 2 means that f(x) divides x . Up to scalar multiples, the only divisors of x are x and x .

(c). R[x]/(x2 3x + 2). − Solution. The nonunit divisors of x2 3x +2=(x 1)(x 2) are, up to scalar multiples, x2 3x + 2 and x 1 and x 2. The− maximal ideals− are− (x 1) and (x 2), since the quotient− of R[x] by− either of those− ideals is the ﬁeld R. − −

(d). R[x]/(x2 + x + 1).

Solution. Since every ideal in R[x] is principal, an ideal is maximal if and only if it is generated by a polynomial that cannot be factored into nonunit factors. x2 + x + 1 cannot be factored since its two degree one factors have nonreal coeﬃcients (nonreal roots). Hence, (x2 + x + 1) is a maximal ideal of the polynomial ring. Hence, the quotient ring of the problem is a ﬁeld, and (0) is the maximal ideal.

3 3 3.(8.3) Prove that F2[x]/(x + x + 1) is a ﬁeld, but that F3[x]/(x + x + 1) is not a ﬁeld.

1 Solution. R/J is a field iff J is a maximal ideal in the commutative ring R, by the correspondence theorm. Let F be a field and F [x] the polynomial ring over F . Every ideal in F [x] is principal. For f(x) F [x], the ideal (f(x)) is maximal iff f(x) is an irreducible polynomial, since for any ideal∈ (g(x)) in F [x], (f(x)) (g(x)) if and only if g(x) divides f(x), i.e., if f(x) is a polynomial multiple of g(x). ⊂ A degree 2 or 3 polynomial f(x) F [x] is irreducible if and only if it has no linear factor x a F [x], i.e., no root in F . ∈ − ∈ 3 3 x + x + 1 is irreducible in F2[x] since it has no root in F2, but x + x + 1 is reducible in 3 3 F3[x] since x = 1 is a root. Hence, F2[x]/(x + x + 1) is a field, but F3[x]/(x + x + 1) is not a field.

4.(8.4) Establish a bijective correspondence between the maximal ideals of R[x] and the points of the upper half plane.

Solution. We know that the maximal ideals in R[x] are generated by the irreducible polynomials The bijective correspondence from the points in the upper half of C to the maximal ideals of R[x] takes a real number a to the maximal ideal (x a) and takes a complex number − 2 2 2 a + bi, b> 0, to the maximal ideal ((x (a + bi))(x (a bi))) = (x 2ax + a + b ). That mapping is injective since those ideals− are all distinct.− We− need to show− that those maximal ideals are all the maximal ideals of R[x], we need to show that every irreducible polynomial in R[x] has degree 1 or 2. To see that, begin with the fact that C is a 2-dimensional real vector space, and the fact that every polynomial in R[x] has its roots in C. Suppose that f(x) R[x] is irreducible and that c C is a root of f(x). Then the homomorphism φ : R[x] ∈ C that evaluates each polynomial∈ g(x) R[x] at c has f(x) in its kernal since f(c) =→ 0. Thus, we get a ′ ∈ homomorphism φ : R[x]/(f(x)) C from the quotient ring. Since f(x) is irreducible in R[x], (f(x)) is a maximal ideal, and→ so, it must be the whole kernel of φ. Hence, φ′ is injective, and so, since C is a 2-dimensional real vector space, the real vector space R[x]/(f(x)) that injects into C has dimension 1 or 2. To conclude, the dimension of R[x]/(f(x)) equals the degree of f(x), and so, f(x) has degree 1 or 2.

Chapter 12.

5.(1.3) (a). Let n and m be relatively prime integers and let a and b be any integers. Show that there is an integer x such that x = a in Z/mZ and x = b in Z/nZ.

2 Solution. Because n and m are relatively prime, Zm Zn = Zmn. Hence, the kernel of the natural homomorphism ∩ Z Z/mZ Z/nZ → × equals Zmn. Therefore, there is the injection

φ′ : Z/mnZ Z/mZ Z/nZ, → × and since the two rings have the same number mn elements, the injection is a bijection, and the two rings are isomorphic. Take any element (a, b) in Z/mZ Z/nZ, and take x to be the corresponding element in × Z/mnZ under the isomorphism.

We also know this result from problem 4 in assignment 5 since (m)+(n)= Z when m and n are relatively prime.

(b). Find all such solutions x′ to part (a).

Solution. For x′, choose any element of the coset x + mnZ.

6.(2.6) Prove that the following rings are Euclidean domains. (a). Z[ω], ω = e2πi/3.

Solution. Here we are following the book’s proof that the Gaussian integers form a Eu- clidean domain. See the discussion of problem 7 below before beginning the solution to problem 6. Deﬁne the size of an element a + bω to be the square of its length as a complex number. The size of ω = 1/2+ i√3/2 is 1. − The fundamental domain for the lattice in C given by the points of Z[ω] is the parallelogram determined by 1 and ω. The four circles with centers at the four vertices of the parallelogram, each passing− through the center of the parallelogram, cover the parallelogram. The long diagonal of the parallelogram goes from 0 to 2ω and has length 2. Drawing a pic- ture, one sees that all points of parallelogram lie with distance 1 of some vertex, and so, at distance-square less than one of some vertex. Therefore, each complex number is at distance-squared less than 1 of some point of the lattice. To divide c + dω by a + bω (elements of Z[ω]), locate (a + bω)−1(c + dω) at distance- squared less than 1 of some point x + yω of the lattice. Since the size measurement is multiplicative, the size

(c + dω) (a + bω)(x + yω) 2 = (a + bω)((a + bω)−1(c + dω) (x + yω)) | − | | − |

3 is less than a + bω 1, as required. | | (b). Z[√ 2]. − Solution. Deﬁne the size of a + b√ 2 to be a2 +2b2. The elements a + b√ 2 form a rectangular lattice in C of side lengths− 1 and √2. Each point in the rectangle− is within distance √3/2 of one of the vertices, and so within distance-squared of 3/4 of one of the vertices. We can repeat the argument in (a) and in 7., since the distance-squared is less than 1 to conclude that the domain is Euclidean.

7.(2.8). Describe a systematic way to do division with remainder in Z[i]. Use it to divide 4+36i by 5+ i.

Solution. The size of a + bi is defined to be a2 + b2, the square of its length as a complex number. We can always divide an element by a unit without remainder. The goal of division of c + di by a nonunit a + bi is to produce an expression c + di (x + yi)(a + bi) that has size less than the size of a + bi, for some values of x and y in Z. − To do that, the elements of the Gaussian integers form a square lattice in C of side length 1. Draw 4 circles with centers at the four vertices, passing through the center of the square (they have radius 1/√2). Those circles cover the square. Hence, each point in the square is within distance 1/√2 of at least one of the vertices, i.e., within distance-squared 1/2. Each complex number lies in one of the squares, and so the minimum distance-squared from a complex number to a point of the lattice is less than or equal to 1/2. To divide c + di by a + bi, locate (a + bi)−1(c + di) in some square, and find the closest vertex x + yi to (a + bi)−1(c + di). Then the squared-distance between those points is no more than 1/2. Multiplying both points x + yi to (a + bi)−1(c + di) by a + bi, the distance-squared from (x + yi)(a + bi) to c + di is no more than a + bi 2/2, division sought in the first paragraph. | | To do that explicitly when we want to divide 4 + 36i by 5+ i, consider

− (5 + i) 1(4+36i) = ((5 i)/26)(4 + 36i)=28/13+ (88/13)i. − That number is located in lattice square with vertics at 2 + 6i,3+6i,2+7i, and 3+7i. Its distance squared to the vertex 2 + 7i is 2/13+(3/13)i 2 1/2. Hence, | − | ≤ 4+36i =(2+7i)(5 + i) + [(4 + 36i) (2+7i)(5 + i)]=(2+7i)(5 + i)+(1 i), − − where 1 i 2 < 5+ i 2. | − | | | 8.(2.9). Let F be a ﬁeld. Prove that the ring F [x, x−1] of Laurent polynomials is a principal ideal domain.

Solution. F [x] is a principal ideal domain.

4 The elements of the ring can be expressed uniquely as f(x)/xn, for some f(x) F [x], n 0, −n ∈ ≥ m and f(x) has nonzero constant term. In fact, If we take an element anx + +a0+ +amx n m+n n with a = 0, then the element equals (a + a −1x + a0x + a x )/x . n n n m To show that an ideal J ′ in F [x, x−1] is principal, consider the intersection J = J ′ F [x] of J ′ with the subring F [x]. Then J is a principal ideal (f(x)) for some monic polynomial∩ f(x), and f(x) generates J ′ in F [x, x−1] too. In fact, if g(x)/xm is any element of J ′, then g(x) is an element of J, and so, g(x)= h(x)f(x) for some h(x) since f(x) generates J. Then g(x)/xm =(h(x)/xm)f(x) is an element of the ideal generated by f(x) in F [x, x−1]. More generally, the same proof shows that if R is a principal ideal domain and S is a multiplicative subset of R, then a/b : a R, b S is also a principal ideal domain. { ∈ ∈ }