Download 1 Principal Ideal Domains

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Principal Ideal Domains
Example. Consider the ring
√
√
R = Z[ −5] = {a + b −5 | a, b ∈ Z}.
Then R does not always have GCDs.
√
√
To see this, √
notice that 6 = 2 · 3 = (1 + −5)(1 − −5). Then it can
√ be shown that no GCD
of 6 and 2(1 + −5) exists. To see this, notice
that
both
2
and
1
+
−5 are common divisors
√
√
of 6. So if d were a GCD of 6 and 2(1 + −5), both
√ 2 and 1 + −5 would have to divide it.
However, taking the complex norm of a number in Z[ −5] still makes sense, and still plays nicely
with multiplication. So,
2|d → 4 divides |d|2 ,
and
(1 +
√
−5)|d → 6 divides |d|2 .
But also,
d|6 → |d|2 divides 36,
√
so these together imply that |d|2 = 12 or 36. But no element of Z[ −5]√can have norm 12, so
|d|2 = 36. Then |6/d|2 = ±1, implying
that d = ±6 (since not much in Z[ −5] has norm 1). But
√
of course ±6 don’t divide 2 + 2 −5 and so can’t be the GCD. We saw before that every Euclidean Domain R has the property that each ideal is generated
by a single element. Let’s give this property a name.
Definition. An integral domain in which every ideal is principal is called a Principal Ideal
Domain (PID).
Now, most of what we said about GCDs in Euclidean Domains applies in PIDs as well.
Theorem 1. Let R be a PID. Let a, b ∈ R be non-zero elements and let d be an element such that
(d) = (a, b). Then:
a.) d is a GCD of a and b.
b.) d = ax + by for some x, y ∈ R.
c.) d is unique up to multiplication by a unit in R
We’ve already proved all of these - the only thing we needed the Euclidean Domain property
for was a way to find that GCD. In PIDs, we at least are guaranteed that the GCD will exist,
although it may in general be hard to find.
So, what’s so great about PIDs? Well, number one is the following.
Theorem 2. Let R be a PID. Then a nonzero ideal I of R is prime if and only if it is maximal.
Proof. We already know that a maximal ideal is necessarily prime. So let P be a non-zero
prime ideal of R. Assume that M is another ideal with P ⊂ M . Since R is a PID, we can write
P = (p) and M = (m). Then (p) ⊆ (m), which means that mr = p for some r ∈ R. Thus either m
or r is in P . By assumption, m ∈
/ P , so that r ∈ P . Then r = sp for some s ∈ R. Substituting,
that means msp = p, so that ms = 1 and m is a unit. Thus M = R. So any ideal bigger than P is
all of R and P is maximal. We’ve talked about how F [x] is a PID for any field F . In fact, a sort of converse is true.
Proposition 3. Let R be a commutative ring so that R[x] is a PID. Then R is a field.
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Proof. Since R[x] is a PID, it has an identity, so R must have an identity as well. Also, R has
no zero divisors since it is a subring of R[x]. Then consider the ideal (x). This ideal is prime in
any polynomial ring R[x], since R[x]/(x) ∼
= R is an integral domain. Since R[x] is a PID, (x) is
maximal, and R[x]/(x) ∼
= R is actually a field. 2
Unique Factorization Domains
We’ve talked about GCDs in general rings. The next nice thing from Z that we’d like to mimic is
unique factorization into products of primes.
Definition. Let R be an integral domain. An element r ∈ R is irreducible if, anytime we can
write r = ab in R, one of a or b is a unit. An element r is called prime if the ideal generated by r
is prime. That is, r is not a unit and r|ab implies r|a or r|b. Two elements a and b are said to be
associates if a = ub for some unit u ∈ R.
Examples. In Z, every prime number is both prime (in this sense) and irreducible.
In R[x], any linear polynomial (x + a) is irreducible and prime. Also, quadratics of the form
2
x + a, where a > 0, are both prime and irreducible.
In general, prime and irreducible want to be the same thing - they both try to capture the idea
that the element can’t
√ be divided any further in the ring, except in trivial ways.
Note that√in Z[ −5],
√ the number 2 is irreducible but not prime. It is not prime because 2
divides (1 + −5)(1 − −5) = 6, but 2 does not divide either of the factors. It is still irreducible
because |2|2 = 4, so if we can write 2 = xy, then either |x|2 = 2 = |y|2 (which is impossible), or
else (WLOG) |x|2 = 1, which implies x = ±1, which are units.