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Mathematics 228(Q1), Assignment 5
Solutions
Exercise 1.(10 marks) Given functions f : T → R and g : T → S, their product f × g : T → R × S is the
function defined by
(f × g)(t) = (f (t), g(t)),
t ∈ T.
Assuming f and g are homomorphisms of rings, show f × g is a homomorphism of rings. Furthermore, show
that f × g is injective if both f and g are injective.
Solution. If t1 and t2 are elements of T then
(f × g)(t1 + t2 ) = (f (t1 + t2 ), g(t1 + t2 ))
= (f (t1 ) + f (t2 ), g(t1 ) + g(t2 ))
= (f (t1 ), g(t1 )) + (f (t2 ), g(t2 ))
= (f × g)(t1 ) + (f × g)(t2 ),
(f and g are hom.)
and
(f × g)(t1 t2 ) = (f (t1 t2 ), g(t1 t2 ))
= (f (t1 )f (t2 ), g(t1 )g(t2 ))
= (f (t1 ), g(t1 )) (f (t2 ), g(t2 ))
= (f × g)(t1 )(f × g)(t2 ).
(f and g are hom.)
Thus, f × g is a homomorphism of rings.
Suppose f and g are both injective. If
(0, 0) = (f × g)(t) = (f (t), g(t))
then
f (t) = 0
and
g(t) = 0.
Since f and g are both injective, we deduce t = 0. It follows that the homomorphism f × g is injective.
√
√
√
Exercise 2.(10 marks) Let Z[ 2] be the ring of exercise 2, assignment 4 and f : Z[ 2] → Z[ 2] the function
defined by
√
√
f (a + b 2) = a − b 2.
√
Show f is an isomorphism of Z[ 2].
Solution. If
√
x1 = a1 + b1 2
and
√
x2 = a2 + b2 2
√
are two elements of Z[ 2] then
√
√ f (x1 + x2 ) = f (a1 + b1 2) + (a2 + b2 2)
√ = f (a1 + a2 ) + (b1 + b2 ) 2
√
= (a1 + a2 ) − (b1 + b2 ) 2
√
√
= (a1 − b1 2) + (a2 − b2 2)
√
√
= f (a1 + b1 2) + f (a2 + b2 2) = f (x1 ) + f (x2 )
and
√ √
f (x1 x2 ) = f (a1 + b1 2)(a2 + b2 2)
√ = f (a1 a2 + 2b1 b2 ) + (a1 b2 + a2 b1 ) 2
√
= (a1 a2 + 2b1 b2 ) − (a1 b2 + a2 b1 ) 2
√
√
= (a1 − b1 2)(a2 − b2 2)
√
√
= f (a1 + b1 2)f (a2 + b2 2) = f (x1 )f (x2 ).
We conclude that f is a homomorphism of rings.
√
√
It remains to check f is bijective. Observe that if x = a + b 2 belongs to Z[ 2] then
√ f ◦ f (x) = f (f (x)) = f f (a + b 2)
√
= f (a − b 2)
√
√
= a − (−b) 2 = a + b 2 = x.
Since x was essentially arbitrary, we deduce
f ◦ f = I,
√
where I is the identity function on Z[ 2]. In particular, f is an invertible function, hence a bijection.
Exercise 3.(10 marks) Let R be the ring of all functions f : R → R. Given a ∈ R, the function a : R → R
defined by
a (f ) = f (a),
f ∈ R.
is referred to as evaluation at a. Verify that a is a homomorphism of rings. Is a injective ? surjective ? Be
sure to justify your answers.
Solution. Let f and g be elements of the ring R. In light of the definition of the sum and product of two
functions, we calculate
a (f + g) = (f + g)(a) = f (a) + g(a) = a (f ) + a (g)
and
a (f g) = (f g)(a) = f (a)g(a) = a (f )a (g).
This shows that a is a homomorphism of rings.
Observing
a (x − a) = a − a = 0
but x − a is not the zero function, we deduce a is not injective. On the other hand, if b ∈ R then the
function
f (x) = x − a + b
has the property
a (f ) = f (a) = a − a + b = b.
It follow that a is surjective.
Exercise 4.(10 marks) Let f : R → S be a homomorphism of rings. Show that if r ∈ R then for all n ∈ Z
f (nr) = nf (r).
(Hint : Recall definition of multiples and the techniques used to prove their elementary properties.)
Solution. We observe
f (0r) = f (0) = 0 = 0f (r).
Assuming
f (kr) = kf (r)
for k ≥ 0, the inductive definition of non-negative multiples yields
f ((k + 1)r) = f (kr + r)
= f (kr) + f (r)
= kf (r) + f (r)
= (k + 1)f (r).
(f is hom.)
(IH)
(PMI) allows us to deduce that if n ∈ N then
f (nr) = nf (r).
If n < 0 then we can write n = −m for some m ∈ N. Since homomorphisms preserve additive inverses, the
definition of negative multiples yields
f (nr) = f ((−m)r) = f (−(mr)) = −f (mr) = − (mf (r)) = (−m)f (r) = nf (r).
Thus the relation
f (nr) = nf (r)
holds for all n ∈ Z.
Exercise 5.(10 marks) Given a homomorphism f : R → S of rings, its kernel is the set
ker f = {r ∈ R : f (r) = 0S }.
(a) Verify that ker f is a subring of R.
(b) Show that ker f enjoys the following additional property : if k ∈ ker f and r ∈ R then both rk and kr
lie in ker f .
Solution.(a) Since f is a homomorphism,
f (0R ) = 0S .
Thus, 0R ∈ ker f ; in particular, ker f is non-empty. If f (r1 ) = f (r2 ) = 0S the elementary properties of
homomorphisms yield
f (r1 + r2 ) = f (r1 ) + f (r2 ) = 0S + 0S = 0S ,
f (r1 r2 ) = f (r1 )f (r2 ) = 0S 0S = 0S , and
f (−r1 ) = −f (r1 ) = −0S = 0S .
In light of the preceding calculation, if r1 and r2 belong to ker f then so do r1 + r2 , r1 r2 and −r1 , i.e. ker f
is closed under addition, multiplication and additive inverses. This completes the verification that ker f is a
subring of R.
(b) By definition of ker f , f (k) = 0S . Therefore,
f (kr) = f (k)f (r) = 0S f (r) = 0S ,
and
f (rk) = f (r)f (k) = f (r)0S = 0S .
The preceding calculations allow us to deduce kr and rk are both elements of ker f .
Exercise 6.(10 marks) Recall, if f : R → S and g : S → T are functions then their composite g ◦ f : R → T
is the function defined by
(g ◦ f )(r) = g(f (r)),
r ∈ R.
(a) Assuming f and g are homomorphisms of rings, show g ◦ f is also a homomorphism.
(b) If f and g are isomorphisms, show g ◦ f is an isomorphism.
Solution.(a) We have to verify that g ◦ f preserves addition and multiplication. Given r1 , r2 ∈ R, we
calculate
g ◦ f (r1 + r2 ) = g(f (r1 + r2 ))
= g(f (r1 ) + f (r2 ))
= g(f (r1 )) + g(f (r2 ))
= (g ◦ f )(r1 ) + (g ◦ f )(r2 ),
( f is a hom.)
( g is a hom.)
g ◦ f (r1 r2 ) = g(f (r1 r2 ))
= g(f (r1 )f (r2 ))
= g(f (r1 ))g(f (r2 ))
= (g ◦ f )(r1 )(g ◦ f )(r2 ).
( f is a hom.)
( g is a hom.)
and
Thus, g ◦ f is a homomorphism.
(b) In light of (a), it is sufficient to show that g ◦ f is a bijection. Suppose
0T = g ◦ f (r) = g(f (r)).
Since g is an isomorphism, g is injective. Therefore, the preceding equation allows us to conclude
f (r) = 0S .
Observing f is injective, being an isomorphism, we deduce r = 0R . The fact g ◦ f is a homomorphism allows
us to conclude that g ◦ f is injective.
Let t ∈ T . Since g is surjective, being an isomorphism, there exists s ∈ S such that g(s) = t. As f is
also an isomorphism, there exists r ∈ R such that f (r) = s. We deduce
g ◦ f (r) = g(f (r)) = g(s) = t.
Since t was essentially arbitrary, we conclude g ◦ f is surjective, hence a bijection.
Alternate Solution. Let f −1 and g −1 be the inverse functions of f and g, and consider the composite
f −1 ◦ g −1 : T → R.
We note
(g ◦ f ) ◦ (f −1 ◦ g −1 ) = g ◦ (f ◦ f −1 ) ◦ g −1 = g ◦ IS ◦ g −1 = g ◦ g −1 = IT ,
and
(f −1 ◦ g −1 ) ◦ (g ◦ f ) = f −1 ◦ (g −1 ◦ g) ◦ g = f −1 ◦ IS ◦ f = f −1 ◦ f = IR .
We deduce f −1 ◦ g −1 is an inverse of g ◦ f , hence g ◦ f is a bijection.
Exercise 7.(10 marks) Let f : R → S be an isomorphism of rings and let g : S → R be the inverse function
of f . Show g is an isomorphism of rings.(Hint : To verify g(s1 + s2 ) = g(s1 ) + g(s2 ), consider its image under
the homomorphism f and use the facts that f is a homomorphism and f ◦ g is the identity map.)
Solution. We have to verify that g preserves both additon and multiplication. Given s1 and s2 in S, the
fact f ◦ g = IS yields
f (g(s1 + s2 )) = (f ◦ g)(s1 + s2 ) = IS (s1 + s2 ) = s1 + s2 .
On the other hand, using the fact f is a homomorphism
f (g(s1 ) + g(s2 )) = f (g(s1 )) + f (g(s2 )) = (f ◦ g)(s1 ) + (f ◦ g)(s2 ) = IS (s1 ) + IS (s2 ) = s1 + s2 .
The preceding calculations show
f (g(s1 + s2 )) = f (g(s1 ) + g(s2 )) .
The fact f is injective allows to conclude
g(s1 + s2 ) = g(s1 ) + g(s2 ),
i.e. g preserves addition.
Similarly, one has
f (g(s1 s2 )) = (f ◦ g)(s1 s2 ) = IS (s1 s2 ) = s1 s2 .
On the other hand, using the fact f is a homomorphism
f (g(s1 )g(s2 )) = f (g(s1 ))f (g(s2 )) = (f ◦ g)(s1 )(f ◦ g)(s2 ) = IS (s1 )IS (s2 ) = s1 s2 .
The preceding calculations show
f (g(s1 s2 )) = f (g(s1 )g(s2 )) .
The fact f is injective allows to conclude
g(s1 s2 ) = g(s1 )g(s2 ),
i.e. g preserves multiplication.
The preceding shows that g is a homomorphism, hence an isomorphism (as g is bijective, with inverse f ).