Download 2 Integral Domains and Fields

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Integral Domains and Fields
Definition 2.1. The characteristic, char(R), of a ring R is the least positive integer n such that 1 + · · · + 1 = 0 (n 1s), that is, such that n · 1 = 0.
If, as may well be the case, there is no such positive n, then the characteristic of R is defined to be 0.
For example, Z has characteristic 0, as have the rings Q and C. On the
other hand char(Zn ) = n.
Lemma 2.2. Suppose that char(R) = n > 0. Then:
(i) n · r = 0 for every r ∈ R;
(ii) if m is a positive integer then m · 1 = 0 iff n|m.
Proof. (i) n · r = r + ... + r = r · (1 + ... + 1) = r × 0 = 0.
(ii) Let m ∈ N with m · 1 = 0 and write m = nq + r where 0 ≤ r ≤ n − 1
and q ≥ 0 are integers. Then
0 = (nq + r) · 1 = q · (n · 1) + r · 1 = 0 + r · 1 = r · 1.
By the minimality of n, we must have r = 0 and so n|m. Conversely if n|m,
then m = nq and so m · 1 = (nq) · 1 = q · (n · 1) = 0.
Definition 2.3. The ring R is a domain if, for all r, s ∈ R
rs = 0 ⇒ r = 0 or s = 0.
A non-zero element r ∈ R is a zero-divisor if there is a non-zero element
s ∈ R with rs = 0 or sr = 0. So a domain is a ring without zero-divisors.
A commutative domain is also called an integral domain. (Recall that
a ring R is said to be commutative if the multiplication is commutative:
rs = sr for all r, s ∈ R.)
Examples 2.4. (1) Z, Q, R and C are domains; (2) Z[i] is a domain - this
can be proved directly or use the fact that C is a domain then apply the next
result.
Lemma 2.5. If S is a subring of R and R is a domain then S is a domain.
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Proof. Let r, s ∈ S with rs = 0. Then rs = 0 in R and so r = 0 or s = 0.
Examples 2.6. the polynomial ring Z[X] is a domain. The next result
proves this in the general case.
Proposition 2.7. Suppose that R is a domain. Then the polynomial ring
R[X] is a domain.
Proof. Let f, g ∈ R[X] with f, g ̸= 0. We can write
f=
n
∑
ai X i ,
g=
i=0
m
∑
bi X i
i=0
where ai , bi ∈ R for all i and an , bm ̸= 0. The coefficient of X n+m in f g is
an bm . Since R is a domain, an bm ̸= 0 and so f g ̸= 0.
Corollary 2.8. Suppose that R is a domain. Then the ring, R[X1 , . . . , Xn ],
of polynomials in n indeterminates and with coefficients in R, is a domain.
Proof. Recall the definition
R[X1 , ..., Xn ] = R[X1 , ..., Xn−1 ][Xn ]
for n > 1. Therefore we can use induction and Prop 2.7 to prove that
R[X1 , ..., Xn ] is a domain.
Definition 2.9. A division ring is a ring in which every non-zero element
has a right inverse and a left inverse: for every r ∈ R there is s ∈ R such
that rs = 1 (r is right invertible) and there is t ∈ R such that tr = 1 (r is
left invertible).
In this case (see below) s = t and we write r−1 for this inverse of r and
say just that r is invertible or that r is a unit.
A field is a commutative division ring.
√
√
Examples 2.10. Q, R, C and Q[ 2] = {a+b 2 | a, b ∈ Q} are all examples
of fields.
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Lemma 2.11. If R is a ring and r ∈ R has both a right and a left inverse
then these are equal.
Proof. Let s be the right inverse and t be the left inverse of r. Consider trs:
trs = (tr)s = 1s = s
trs = t(rs) = t1 = t.
So s = t.
Lemma 2.12. The following conditions on the integer n ≥ 2 are equivalent:
(i) Zn is an integral domain;
(ii) Zn is a field;
(iii) n is a prime.
Proof. (i)⇒(ii): Suppose Zn is a domain. Let r ∈ Zn with r ̸= 0. Consider
Ar = {0r, 1r, 2r, ..., (n − 1)r}. Since Zn is a domain, the elements of Ar are
distinct because
ir = jr ⇒ (i − j)r = 0 ⇒ i − j = 0 ⇒ i = j.
Therefore Ar = Zn and there is some element of Ar equal to 1. So sr = 1 for
some s ∈ Zn . Since Zn is commutative, r is invertible and therefore Zn is a
field.
(ii)⇒(iii): Suppose that Zn is a field and assume n is not a prime. Then
n = mq where 1 < m, q < n and m, q ∈ Z. In Zn we have mq = 0 with
m, q ̸= 0. Since Zn is a field, m−1 exists and
0 = m−1 mq = 1q = q
a contradiction. So n is prime.
(iii)⇒(i): Suppose n is a prime and let r, s ∈ Zn with rs = 0. Then n|rs
and so n|r or n|s. Therefore r = 0 or s = 0 in Zn . Hence Zn has no zero
divisors.
Beware that an element of a non-commutative ring can have, for instance,
a left inverse without having a right inverse and, in such a case, there might
be more than one left inverse.
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Example 2.13. Let K be your favourite field and let V be the (infinitedimensional) vector space over K with basis v1 , v2 , . . . , vn , . . . . Let R be the
ring of all linear transformations from V to itself, the operations on R being
(pointwise) addition of linear transformations and, for the “multiplication”
on R, use composition of linear transformations (with the convention that sr
means “do r then do s”).
Note that an invertible element of R must be a bijection from V to itself.
Let r ∈ R be “right-shift”: the linear map which is defined on the given
basis by sending vn to vn+1 . Since this map is not surjective, it is not an
isomorphism, hence r is not invertible.
But r is left invertible: if s is the linear map which sends vn+1 back to
vn and sends v1 to 0, then sr = 1 (where 1 is, note, the identity map on V ).
Let t be the linear map which sends vn+1 to vn and sends v1 to v1 : then also
tr = 1, showing that a left inverse need not be unique.
Proposition 2.14. Every division ring is a domain. Every field is an integral
domain.
Proof. Let R be a division ring and suppose r, s ∈ R with rs = 0. Assume
r ̸= 0. Then r−1 exists and
0 = r−1 (rs) = (r−1 r)s = 1s = s.
Therefore R contains no divisors of zero.
Example 2.15. There are division rings which are not fields, that is, which
are not commutative. The best known is the ring of quaternions: H =
{a + bi + cj + dk | a, b, c, d ∈ R} where i2 = j 2 = k 2 = −1 and ij = k, jk = i,
ki = j. From these relations it follows, for example, that ji = −k ̸= k = ij.
Notice that this contains, as a subring, the field, C, of complex numbers.
And, in case you were wondering, “H” stands for “Hamilton”. It can be
shown that every non-zero element has an inverse.
Lemma 2.16. In any ring R, the set of units R∗ forms a a group under
multiplication.
Proof. Let a, b ∈ R∗ be units ie. they have inverses a−1 and b−1 . Then the
product ab is also a unit with inverse b−1 a−1 because
(ab)(b−1 a−1 ) = a(bb−1 )a−1 = aa−1 = 1
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and similarly (b−1 a−1 )(ab) = 1. Multiplication in R∗ is associative because
R is a ring. The identity is 1 (1 is invertible with 1−1 = 1) and by definition
each element in R∗ has an inverse. Hence R∗ is a group.
Definition 2.17. An element r of a ring R is nilpotent if there is some
integer n ≥ 1 with rn = 0 (and the least such n is the index of nilpotence
of r).
An element r ∈ R is idempotent if r2 = r. For example 0 and 1 are
idempotent.
We finish this section with some further examples of rings:
Examples 2.18. (1) Here is a general construction: if R1 and R2 are rings
then their product is their set-theoretic product R1 × R2 made into
a ring by defining the operations + and × (don’t confuse “×” and
“×”!) by (r1 , r2 ) + (s1 , s2 ) = (r1 + s1 , r2 + s2 ) and (r1 , r2 ) × (s1 , s2 ) =
(r1 × s1 , r2 × s2 ) (to be more precise we could write (r1 +1 s1 , r2 +2 s2 )
to show that the additions refer to operations in different rings and
similarly for multiplication). The element (0, 0) clearly is a, hence the,
zero of this ring and the element (1, 1) is the identity element of R1 ×R2 .
Note that the elements (1, 0) and (0, 1) corresponding to the identity
elements of the original rings are idempotents of the ring R1 × R2
and they are orthogonal in the sense that idempotents e1 and e2 are
orthogonal if e1 e2 = 0 = e2 e1 .
(2) Consider the ring, R[X], of polynomials in X with coefficients from R.
Now extend this by a new variable Y which does not commute with X
but, rather, satisfies the relation Y X = XY + 1. We denote this ring,
called the first Weyl algebra, by R⟨X, Y : Y X = XY +1⟩ (thus, given
R, we are describing this ring in terms of generators and relations): its
elements are “non-commutative polynomials” in X and Y. Every such
polynomial can be brought into a “normal form”: a sum of monomials,
each of which has the form X m Y n for some non-negative integers m, n.
(3) Let R be any ring and let G ∑
be any group. The group ring RG is
the set of formal finite sums
ri gi (up to the usual identifications)
where ri ∈ R and gi ∈ G. Addition is formal addition using the rule
rg +sg = (r +s)g and multiplication is based on (rg)×(sh) = (rs)(gh).
The one in RG is 1e where 1 is the one of R and e is the identity of G.
The zero of RG is the element 0e.
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