Download 8. Check that I ∩ J contains 0, is closed under addition and is closed

8. Check that I ∩ J contains 0, is closed under addition and is closed
under multiplication by elements in the ring, using that both I and J are.
(For instance 0 ∈ I, 0 ∈ J =⇒ 0 ∈ I ∩ J etc.)
9. Every ideal contains zero. If I = {0} (no other elements), then I = 0Z,
in other words we take n = 0. Otherwise there is some non-zero integer m
in it, and −m = m × (−1) as well. One of m, −m is positive, so I contains
at least one positive number. Every non-empty subset of N has a smallest
element, so let n be the smallest member of I ∩ N.
We claim that I = nZ. One inclusion is clear: n ∈ I =⇒ nZ ⊂ I,
because I is closed under multiplication with elements of Z. Suppose I
contains some other number k ∈
/ nZ. Using division with remainder, write
k = na + b with 0 < b < n. Then k ∈ I, na ∈ I =⇒ b ∈ I, contradicting
the choice of n as the smallest positive element of I.
10. (1) If φ : Z → R is a homomorphism, recall that φ(1) = 1 and
φ(0) = 0. Because φ preserves addition, φ(1 + . . . + 1) (n times) must be
1 + . . . + 1 (n times) in R. In other words, φ is uniquely determined on
all non-negative integers — no choice here. Because φ preserves ‘-’ as well
(φ(−n) = −φ(n)), it is uniquely determined on the negative integers as well.
So it is uniquely determined on the whole of Z. This shows uniqueness.
For the existence, define φ as above. In other words, let φ(0) = 0, φ(1) =
1, φ(1 + . . . + 1) = 1 + . . . + 1 for positive integers and φ(−n) = −φ(n) for
negative ones. Now (this is slightly tedious) prove that φ preserves addition
and multiplication, using the fact that Z and R are rings.
(2) By the Fundamental Homomorphism Theorem, im φ ∼
= Z/ ker φ.
11. This was done in the lectures for m = 2, n = 3, and the argument
for any coprime m and n is the same: define φ : Z → Z/mZ × Z/nZ by
φ(a) = (a mod m, a mod n). This is a homomorphism, and its kernel
ker φ (=preimage of (0, 0)) consists of integers a for which m|a and n|a.
Because m and n are coprime, this is equivalent to mn|a, so ker φ = mnZ.
By the Fundamental Homomorphism Theorem, Z/mnZ = Z/ ker φ ∼
= im φ.
In particular, | im φ| = mn. But |Z/mZ × Z/nZ = mn| as well, so φ must
be surjective, and we get the required isomorphism. [Note that surjectivity
means that for every pair (b mod m, c mod n) there exists an integer a
(unique up to multiples of mn, in fact), which reduces to b mod m and to c
mod n. This is exactly Chinese Remainder Theorem.]
1
2
12. For example, take m = n = 2. In this case, you get two rings that
are not even isomorphic as groups under +.
13. First, look at the sizes: Z/5Z and Z/7Z have order (=size) 5 and 7,
respectively, and all the others are infinite sets. So these two cannot be in
bijection either with each other or with one of the others. Of the other four,
R × R is the only one that has zero divisors — elements a 6= 0, b 6= 0 for
which ab = 0 (e.g. a = (1, 0), b = (0, 1)). So it cannot be isomorphic to Z, Q
or R, that have no zero divisors. Of these remaining three, Z is not a field,
while the other two are, and R has solutions to an equation x2 = 2 while Q
has not. (Or, come to think of it, Q is countable and R is uncountable.)
14. In Z/5Z we have 1 + 1 + 1 + 1 + 1 = 0 and a homomorphism
φ : Z/5Z → any ring R takes 0 to 0, 1 to 1 and addition to addition, so
1 + 1 + 1 + 1 + 1 = φ(1 + 1 + 1 + 1 + 1) = φ(0) = 0
in R. This means that there are no homomorphisms from Z/5Z to a ring in
which 1 + 1 + 1 + 1 + 1 6= 0 (for instance to R, Q, Z[x], Z/7Z etc.)
For R → Z/5Z, note that R is a field, and every ideal of a field is either
0 or the whole of R (as it must contain a unit if it is non-zero). So the
only homomorphisms R → any ring are either a map to the zero ring or
injections. But clearly R cannot inject into Z/5Z.
The same argument in principle applies to C → R × R (as C is a field), as
long as we can prove that there are no injective homomorphisms C ,→ R×R.
But this is not entirely trivial, because C and R×R have the same cardinality,
so there are set injections from one to the other. So, again, we look at
elements of C that√satisfy an algebraic equation that cannot be satisfied in
R × R. Here i = −1 comes to mind. It satisfies x2 = −1, so its image
y = φ(x) would satisfy y 2 = −1 under a homomorphism φ : C → R × R. But
in R × R every element is of the form (a, b) and its square (a2 , b2 ) cannot be
(−1, −1) because both components are non-negative.