MATH20212 Algebraic Structures 2

Solutions to Example Sheet 4

1. First note that, for any ring R, the identity function on R is an iso-morphism from R to itself and so we have reflexivity of isomorphism.

For symmetry we have to show that if R is isomorphic to S, say θ :R → S is an isomorphism then there is an isomorphism from S toR. Consider the inverse, θ−1 : S → R, of θ, defined by θ−1(s) = r iffs = θ(r). This is a bijection and an isomorphism because, if s, s′ ∈ Swith θ−1(s) = r and θ−1(s′) = r′, then θ(r + r′) = s + s′ and soθ−1(s+ s′) = r + r′ and also θ(rr′) = ss′ and so θ−1(ss′) = rr′.

Next assume that R is isomorphic to S, say θ does the job, and S isisomorphic to the ring T , say ψ : S → T is an isomorphism. ThenR is isomorphic to T because composition, ψθ : R → T , defined by(ψθ)(r) = ψ(θ(r)) is an isomorphism by Lemma 3.10(i) and propertiesof bijections. So isomorphism is transitive and therefore is an equiva-lence relation on the set of rings.

2. LetR denote the ring of matrices of the form a

(1 00 1

)+b

(0 −11 0

)+

c

(i 00 −i

)+ d

(0 ii 0

), with a, b, c, d ∈ R (and i a square root of

−1).

Define the map θ : H → R by

θ(a+ bi+ cj + dk) =

(a+ ci −b+ dib+ di a− ci

).

It is clear that this is 1-1 and onto and that θ(1) is the identity matrix.

Take a+ bi+ cj+ dk, a′+ b′i+ c′j+ d′k ∈ H. To check that θ preservesaddition we have to show that θ

((a+bi+cj+dk)+(a′+b′i+c′j+d′k)

)=

θ(a+ bi+ cj + dk) + θ(a′ + b′i+ c′j + d′k), which is, when you write itdown, obvious.

To check multiplication we have to show that

θ((a+bi+cj+dk)×(a′+b′i+c′j+d′k)

)= θ(a+bi+cj+dk)×θ(a′+b′i+c′j+d′k).

The LHS equals θ((aa′− bb′− cc′−dd′)+ (ab′+a′b+ cd′− c′d)i+(ac′+a′c+ b′d− bd′)j + (ad′ + a′d+ bc′ − b′c)k) = θ(α + βi+ γj + δk), say,

1

=

(α + γi −β + δiβ + δi α− γi

).

The RHS equals

(a+ ci −b+ dib+ di a− ci

)×(a+ ci −b+ dib+ di a− ci

)which, if

you multiply it out, gives exactly the matrix above.

3. Suppose that R is a ring with three elements. Two of these elementsmust be 0, 1; denote the third by a. It will be enough to show thatthere is no choice in drawing up the addition and multiplication tablesof R, for then, if S is another ring with three elements, 0, 1, b say,then the map taking 0 to 0, 1 to 1 and a to b must preserve additionand multiplication (since there is no choice over these), so must be anisomorphism.

Addition: what is 1 + 1? It can’t be 1 because from 1 + 1 = 1 we’dget the contradiction 1 = 0. Could 1 + 1 = 0? No, because by thekind of cancelling argument just used, a+ 1 cannot equal a or 1 hencea+ 1 = 0, but then we’d have 1 + 1 = a+ 1 so a = 1, a contradiction.So we must have 1 + 1 = a, and we’ve also noted that a + 1 = 0, thatis 1 + 1 + 1 = 0 hence the ring is of characteristic 3. It follows thata+ a = 2 + 2 = 1. So now we have the complete addition table.

Multiplication: since 0 times anything is 0 and 1 times anything is thatthing, the only question to be answered is what is a×a? But a = 1+1and so a× a = (1 + 1)a = a+ a = 1.

There was no choice anywhere, so we have proved that, up to isomor-phism, there is just one ring with three elements (familiar as Z3, theintegers modulo 3).

4. We first show that the projection map π1 : R1 × R2 :−→ R1 given byπ((r1, r2)) = r1 is a homomorphism.

Let (r1, r2), (s1, s2) ∈ R1 ×R2. Then

π1((r1, r2) + (s1, s2)) = r1 + s1 = π1((r1, r2)) + π1((s1, s2)),

π1((r1, r2)(s1, s2)) = r1s1 = π1((r1, r2))π1((s1, s2))

and π1((1, 1)) = 1. Therefore π1 is a homomorphism.

The kernel of π1 is {0} ×R2 and therefore π1 is not injective and so isnot an isomorphism.

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5. For the first part, we have already shown that θ is well-defined in thelectures. Let [a]nk, [b]nk ∈ Znk. Then

θ([a]nk+[b]nk) = θ([a+ b]nk) = [a+ b]n = [a]n+[b]n = θ([a]nk)+θ([b]nk)

using properties of modulo arithmetic.

Similarly θ([a]nk[b]nk) = θ([a]nk)θ([b]nk). Also θ([1]nk) = [1]n and so θis a ring homomorphism.

Now suppose there is a homomorphism θ : Zn −→ Zm where m,n ≥ 2.Since n · [1]n = 0 we must have θ(n · [1]n) = 0, that is, n · [1]m = 0. Butthe characteristic of Zm is m and so, by Lemma 2.2(ii), it must be thatm divides n, as required.

6. There are many possible examples. For instance take θ : Q −→ Q[X]to be the embedding which takes a rational r ∈ Q to that same ratio-nal, regarded as a constant polynomial. Take β : Q[X] −→ Q to beevaluation (of a polynomial) at, say 0.

The composition simply returns r ∈ Q to r, so certainly is an embed-ding. But β is not: it takes any polynomial, such as X, with zeroconstant term to 0.

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