Download Math 345 Sp 07 Day 8 1. Definition of unit: In ring R, an element a is

Math 345 Sp 07
Day 8
1. Definition of unit: In ring R, an element a is a unit if it has a multiplicative inverse.
2. Definitions of prime and irreducible. Let R be an I.D.
a. An element p ∈ R is prime if it is not a unit and whenever p|ab for any a,b ∈ R then
either p|a or p|b.
b. An element p ∈ R is irreducible if whenever p=ab for any a,b ∈ R then either a or b is
a unit in R.
Now we can address some questions students had on Day 7:
Question one: In a field are irreducible elements prime? The answer is yes, sort of. Note in field
for any nonzero element p, p|ab ⇒ p|a or p|b is true since px = a and px = b always have a
solution. Also for any nonzero element p, whenever p=ab for any a,b ∈ R then BOTH a and b
are units in R. So everything would be considered prime by these conditions. But just as we don’t
count 1 as prime or irreducible in Z (note it is one of only two units in Z), we don’t count units as
prime or irreducible in an I.D. So in a field, irreducibles are prime and vice versa for the simple
reason that there are none of either in a field.
Question two: Is -2 a prime in 2Z. Answer no, since 2Z is not an I.D. the concept of prime does
not exist in 2Z. -2 is a prime in Z, but not an interesting one since it is just a unit times 2 (so as
far as divisibility goes it is the same as 2).
3. A quick review of quotient groups (this time in a more traditional – less exploratory – order:
a. Lemma: Let H be a subgroup of a group G. Let a, b ∈ G. Prove that aH = bH if and only
if b-1a ∈H. (This is a handy tool for proving that cosets are equal.)
Proof: Let a, b ∈ G.
First, we suppose that aH = bH. We note that clearly a ∈aH. But then since aH = bH we
also have a ∈bH. This means that a = bh for some h∈H. Then b-1a = h. Thus, b-1a∈H.
Second, we suppose that b-1a ∈H. We note that the inverse of this must also be in H since
H is a subgroup. Thus (b-1a)-1 = a-1b∈H as well.
Now suppose x∈aH. Then x = ah for some h∈H. Also since b-1a ∈H, b-1a = h1 for
some h1 ∈H. Thus, a = bh1. Substituting, we get x = bh1h which is clearly in bH.
Similarly, if we suppose x∈bH. Then x = bh for some h∈H. Also since a1b ∈H, a-1b =
h2 for some h2∈H. Thus, b = ah2. Substituting, we get x = ah2h which is clearly in aH.
Thus aH = bH.
So we conclude that aH = bH if and only if b-1a ∈H.
b. Definition. A subgroup H of a group G is normal if ghg-1 ∈H for every g∈ G and h∈
H.
c. Theorem. G/H forms a group if and only if H is a normal subgroup of G (i.e. ghg-1
∈H for every g∈ G and h∈ H).
Proof:
First, suppose that G/H forms a group under coset multiplication.
Let g∈ G and h∈ H. Note that hH = eH where e is the identity in G. Since coset
multiplication is well defined, we must get the same answer when we multiply
these two versions of this coset by gH. So have hgH = gH.
Now by our lemma above, we conclude that ghg-1 ∈H.
Now suppose that H is a normal subgroup of G.
Let a, b, c, d in G be such that aH = cH and bH = dH.
We need to show that aH bH = cH dH. Thus (by applying the operation) we need
to show abH = cdH. By our lemma above, it is sufficient to show that (cd)-1ab ∈
H. Also by our lemma, we note that c-1a and d-1b are both elements of H (since
aH = cH and bH = dH). Let c-1a= h1 and d-1b= h2.
Now, (cd)-1ab = d-1c-1ab. Substituting, we get (cd)-1ab = d-1h1b. Since d-1b= h2,
we have b= dh2. Substituting again, we get (cd)-1ab = d-1h1dh2. Since H is normal,
we know that d-1h1d is an element of H, say d-1h1d = h3. With one last
substitution, we have (cd)-1ab = h3h2 so (cd)-1ab is an element of H. This is what
we needed to show. So we conclude that aH bH = cH dH. So the coset operation
is well defined.
Proving that G/H is a group is then trivial. By definition aH bH = abH which is
clearly a coset (by closure of G) so G/H is closed under the operation. H = eH is
clearly an identity element. Given any coset gH, g-1H is clearly its inverse. Finally
it is a routine exercise to show that this operation is associative.
4. Now for quotient rings! Let R be a ring and S a subring of R. Define R/S = {a + S: a ∈ R}.
The operations on R/S are:
Addition: (a + S) + (b + S) = (a + b) + S. This one is guaranteed to be well defined
since our ring is an abelian group under addition!
Multiplication: (a + S) × (b + S) = (ab) + S. This operation may or may not be well
defined. We need to figure out a necessary and sufficient condition for this to work.
5. Note that R/Z does not form a quotient ring. (Note R is the real numbers here) Note that 1 +
Z = 0 + Z. Now, for any non-integer a, we get: (a + Z)(1 + Z) = a + Z. Since a is not an
integer a + Z is not equal to Z. However, (a + Z)(0 + Z) = 0 + Z = 0. So the multiplication is
not well defined.
6. Prove that addition and multiplication are well defined on Z/4Z. Hint: Convert the lemma
above to additive notation!