Isomorphic to the Product Ring Z/(2) × Z/(3)?

Math 403 Assignment 5. Due Friday, Feb 22, 2013.

1(6.2) (a). Is Z/(6) isomorphic to the product ring Z/(2) × Z/(3)?

Solution. Take the natural homomorphism Z → Z/(2) × Z/(3) that take each integer n to (n+(2), n+(3)). The kernel equals (2)∩(3) = (6). Hence, there is an injective homomorphism Z/(6) → Z/(2) × Z/(3). Since each ring has 6 elements, that homomorphism is surjective; therefore, it is an isomorphism.

(b). Is Z/(8) isomorphic to the product ring Z/(2) × Z/(4)?

Solution. The rings are not isomorphic since in the ring Z/(2)×Z/(4), 4(n+(2), m+(4)) = (0, 0), but in the ring Z/(8), 4(1 + (8)) = 0. Alternatively, since 1 goes to 1 under a ring homomoprhism, there is only one ring homomorphism φ : Z → Z/(2) × Z/(4), φ(n)=(n + (2), n + (4)). The kernel of that homomorphism is (4), not (8). Hence, the rings Z/(8) and Z/(2) × Z/(4) are not isomorphic.

2(6.5) Suppose that we adjoin an element α satisfying the relation α2 = 1 to the ring R of real numbers. Show that the resulting ring is isomorphic to R × R.

Solution. Adjoining an element α satisfying the relation α2 = 1 is the same as forming the ring R[x]/(x2 − 1) = R[x]/((x − 1)(x + 1)). Deﬁne a homomorphism

φ : R[x] → R[x]/(x − 1) × R[x]/(x +1)= R × R, where φ(f(x)) = (f(x)+(x − 1), f(x)+(x + 1)). The kernel is (x − 1) ∩ (x +1) = ((x − 1)(x +1)) = (x2 − 1). Thus, we have an injection R[x]/(x2 − 1) → R × R. That injection is an isomorphism then since each ring is a two-dimensional real vector space. Or you can show directly that it is surjective.

3(6.7) In the ring Z[x], show that (a). (2) ∩ (x)=(2x).

Solution. (2) ∩ (x) consists of the polynomials f(x) with even coeﬃcients and 0 constant term, i.e., f(x)=2xg(x) where we have factored out 2 and x from each term of f(x).

1 (b). Z[x]/(2x) is isomorphic to the subring of Z[x]/(2)×Z[x]/(x)=(Z/(2))[x]×Z of elements (f(x), n) such that f(0) = n in Z/(2).

Solution. Deﬁne a homomorphism Z[x] → (Z[x]/(2) × Z[x]/(x) by mapping f(x) ∈ Z[x] to (f(x)+(2), f(x)+(x) in Z[x]/(2)×Z[x]/(x). The kernel of that homomorphism is (2)∩(x)= (2x). Hence, we have an injection Z[x]/(2x) → (Z[x]/(2) × Z[x]/(x)=(Z/2Z)[x] × Z that takes f(x)to(f(x), f(0)) in (Z/2Z)[x]×Z. Those are the elements (f(x), n)of(Z/2Z)[x]×Z such that f(0) = n in Z/(2), i.e., f(0) and n are either both even or both odd integers. That is why the subring is not the whole target ring.

4(6.8) Let I and J be ideals in a (commutative) ring R such that I + J = R.

(a). Show that IJ = I ∩ J.

Solution. IJ ⊂ I ∩J since I and J are ideals. To show that I ∩J ⊂ IJ when I +J = R, we write 1 = i + j for some i ∈ I and j ∈ J. Then for any element a ∈ I ∩ J, a = a1= ai + aj since ai ∈ IJ and aj ∈ JI = IJ.

(b). Show that R → (R/I) × (R/J) is surjective with kernel IJ = I ∩ J, so that R/IJ is isomorphic to (R/I) × (R/J), when I + J = R.

Solution. Every element of R/I has the form j + I for some j ∈ J since I + J = R, and every element of R/J has the form i + J for some i ∈ I since I + J = R. The element i + j of R is mapped to the general element (j + I, i + J) of (R/I) × (R/J); hence, the homomorphism R → (R/I) × (R/J) is surjective. The kernel of the homomorphism is the intersection I ∩ J of the kernel of the homomorphisms R → R/I and R → R/J.

5(7.1) Show that a domain R of ﬁnite order is a ﬁeld.

Solution. We look for the multiplicative inverse of a nonzero element a ∈ R. If R has n elements, then there must be duplications in the sequence 1, a, a2, ..., an of n + 1 elements. Let ai = aj with 0 ≤ i < j ≤ n. Then ai(1 − ai−j) = 0. Since we are in a domain, ai is nonzero, which forces the factor 1 − aj−i to be 0, i.e., aj−i = 1, and a−1 = aj−i−1.

6(7.3) Can a domain R have exactly 15 elements?

Solution. Because R is a ﬁnite domain, it has prime characteristic p for some prime p. Thus, R contains the ﬁeld Fp of p elements. R is then an Fp vector space, of some dimension m. Then R has pm elements. Since 15 is not a power of a prime, a domain R cannot have 15 elements.

2 7(7.4) Show that the field of fractions of the power series ring F [[x]] over a field F is obtained by inverting the element x. Find a neat description of that field.

∞ n Solution. Consider the set of series {Σn=−manx } where m ranges over the non-negative integers. That set forms a ring under addition and multiplication of series. We want to show that the ring is the ﬁeld of fractions of the power series ring F [[x]] In the 7th problem of assignment 2, we saw that the elements of F [[x]] have the form m 2 n 2 n x (a0 + a1x + a2x + anx + ), with a0 = 0, so that a0 + a1x + a2x + anx + is a unit. The inverse of that element in the ﬁeld of fractions is then

−1 m 2 n −1 −1 m 2 n (x ) (a0 + a1x + a2x + anx + ) =(x ) (b0 + b1x + b2x + bnx + )=

−m −(m−1) n b0x + b1x + + bm + bm+1x + + bm+nx + , 2 n 2 n where b0 + b1x + b2x + bnx + is the inverse of a0 + a1x + a2x + anx + in ∞ n F [[x]]. Therefore the ring of series {Σn=−manx } is the ﬁeld of fractions of F [[x]].

8(7.5) A multiplicative set S in a domain R is a subset that does not contain 0 and is closed under multiplication. Consider the set {a/b|a ∈ R, b ∈ S} in the ﬁeld of fractions of R. Show that that set is a subring of the ﬁeld of fractions.

Solution. We need to show that if b, d ∈ S and a, c ∈ R, then a/b + c/d and (a/b)(c/d) lie in the set {a/b|a ∈ R, b ∈ S}. We have a/b + c/d =(ad + bc)/bd and (a/b)(c/d)=(ac)/bd, which have denominator bd which lies in S since S is multiplicative.