MATH 235 ALGEBRA I
SOLUTIONS TO ASSIGNMENT 7
Page 84 Exercise 5
In the following questions it is useful to remember that under our definitions, a ring homomor
phism takes 1 to 1.
(a) Prove there is no ring homomorphism
Z
/
5
Z
→
Z
.
Assume there exists a ring homomorphism
f
:
Z
/
5
Z
→
Z
. By the definition,
f
(1
Z
/
5
Z
) = 1
Z
But 1
Z
/
5
Z
has additive order 5 and 1
Z
has infinite additive order. Thus
0
Z
=
f
(0
Z
/
5
Z
)
=
f
(1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
)
=
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
)
=
1
Z
+ 1
Z
+ 1
Z
+ 1
Z
+ 1
Z
=
5
Z
Which is a contradiction since 0
Z
= 5
Z
. Thus no such
f
can exist.
(b) Prove that there is no ring homomorphism
Z
/
5
Z
→
Z
/
7
Z
.
Assume there exists a ring homomorphism
f
:
Z
/
5
Z
→
Z
/
7
Z
. By the definition,
f
(1
Z
/
5
Z
) = 1
Z
/
7
Z
But 1
Z
/
5
Z
has additive order 5 and 1
Z
/
7
Z
has additive order 7. Thus
0
Z
/
5
Z
=
f
(0
Z
/
5
Z
)
=
f
(1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
+ 1
Z
/
5
Z
)
=
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
) +
f
(1
Z
/
5
Z
)
=
1
Z
/
7
Z
+ 1
Z
/
7
Z
+ 1
Z
/
7
Z
+ 1
Z
/
7
Z
+ 1
Z
/
7
Z
=
5
Z
/
7
Z
Which is a contradiction since 0
Z
/
7
Z
= 5
Z
/
7
Z
. Thus no such
f
can exist.
(c) Prove that the rings
Z
/
2
Z
×
Z
/
2
Z
and
Z
/
4
Z
are not isomorphic.
Assume there is an isomorphism
f
:
Z
/
4
Z
→
Z
/
2
Z
×
Z
/
2
Z
. We know that
f
(0
Z
/
4
Z
) = 0
Z
/
2
Z
×
Z
/
2
Z
But I also claim that
f
(2
Z
/
4
Z
) = 0
Z
/
2
Z
×
Z
/
2
Z
as well which would mean that
f
is not
injective. To show this, we know that
f
takes 1 to 1, and that every element in
Z
/
2
Z
×
Z
/
2
Z
has additive order 2 (i.e. when you add any element in this ring to itself you get
i
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the additive identity 0). Thus we have
f
(2
Z
/
4
Z
)
=
f
(1
Z
/
4
Z
+ 1
Z
/
4
Z
)
=
f
(1
Z
/
4
Z
) +
f
(1
Z
/
4
Z
)
=
1
Z
/
2
Z
×
Z
/
2
Z
+ 1
Z
/
2
Z
×
Z
/
2
Z
=
(1
,
1) + (1
,
1)
=
(0
,
0)
=
0
Z
/
2
Z
×
Z
/
2
Z
Thus
f
is not injective and hence cannot be an isomorphism.
(d) Is there a ring homomorphism
Z
/
4
Z
→
Z
/
2
Z
×
Z
/
2
Z
?
Since
Z
/
4
Z
is generated by 1, and
f
respects addition, any such homomorphism
f
is
determined by its value at 1. But by our definition,
f
takes 1 to 1. So that
f
(1
Z
/
4
Z
) =
1
Z
/
2
Z
×
Z
/
2
Z
.
Using this
f
is determined as follows.
I will stop using the cumbersome
subscripts, and the ring that each element is in is clear from the context.
f
(1)
=
(1
,
1)
f
(2) =
f
(1 + 1) =
f
(1) +
f
(1)
=
(1
,
1) + (1
,
1) = (0
,
0)
f
(3) =
f
(1 + 2) =
f
(1) +
f
(2)
=
(1
,
1) + (0
,
0) = (1
,
1)
f
(0) =
f
(1 + 3) =
f
(1) +
f
(3)
=
(1
,
1) + (1
,
1) = (0
,
0)
Each calculation is based on the previous one, and it is in this sense that
f
is “determined”
by its value at 1. It is important to note that we have not yet shown
f
is a homomorphism
of rings for we have not shown that
f
respects the operation of multiplication defined
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 Winter '09
 NOONAN
 Homomorphism, ring homomorphism, Epimorphism

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