distributive The law for the field of fractions
Lemma 9.1
a c Suppose that b , d , e F . Then f a b c d
The field of fractions
Theorem 9.2
Suppose that R = 0 is a commutative ring with no zero divisors and let a F = { b : a, b R with b = 0 }. Then F is field with operations a c ad+bc a c ac and b d = bd . b + d = bd Proof We need to show that F is a commutative ring with one and that every non-zero element of F has an inverse. By Lemma 8.2, addition and multiplication are well-defined operations on F . By Lemma 8.3, (F , +) is an abelian group. By Lemma 8.5, the multiplication on F is both associative and commutative and 1F = y is an identity element in F , for 0 = y R. y By Lemma 9.1, the distributive laws hold in F . Hence, F is a commutative ring with one!
+
e f
=
ac bd
+
ae b f
and
a b
+
c d
e f
=
ae b f
+
c e d f
Proof Applying the definitions, e a c a = b cf +de = b d + f df
a(cf +de) b(df ) 8.4 acf b+adeb = = (ac)(bf )+(bd)(ae) bdf b (bd)(bf ) ac a ac = bd + ae = b d + b e f bf
That
a b
+
c d
e f
=
ae b f
+
c e d f
follows by commutativity.
We are now ready to prove that F is a field!
Andrew Mathas (University of Sydney)
Rings, Fields and Galois Theory
Math3962 -- Lecture 9
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Andrew Mathas (University of Sydney)
Rings, Fields and Galois Theory
Math3962 -- Lecture 9
2/9
The field of fractions.../2
Suppose that
a b
Embedding a ring in its field of fractions
a b
= 0F in F . We show that
has an inverse.
Proposition 9.4
Suppose that R = 0 is a commutative ring with no zero divisors, fix a non-zero element x R and let F be the field of fractions of R. Then the map : R - F ; r rx is injective. Hence, F contains a subring x naturally isomorphic to R. Proof Suppose that r , s, R. respects addition 8.4 (r + s) = (r +s)x = rx + sx = (r ) + (s) x x x respects multiplication 8.4 (rs) = (rs)x = (rx)(sx) = rx sx = (r )(s) x xx x x
a 0 That is, b = z , for any 0 = z R (by Lemma 8.3) = az = b 0 = 0 whenever z = 0 = a = 0 since R has no zero divisors = b F a a Therefore, b b = ab = 1F by Lemma 8.5 since ab = 0. a ba
Hence, every nonzero element of F has a multiplicative inverse. Finally, note that 0F = 1F since if y , z = 0 then
0 z
=
y y
since yz = 0.
Definition 9.3
The field F = F (R) in Theorem 9.2 is the field of fractions of R. Example If R = Z then it satisfies the assumptions of Theorem 9.2 since Z is an integral domain = F (Z) Q by construction! = If R = 2Z then the assumptions are still met = F (2Z) Q. =
Andrew Mathas (University of Sydney) Rings, Fields and Galois Theory Math3962 -- Lecture 9 3/9
is injective Suppose that (r ) = (s) = rx = sx x x = (rx)x = (sx)x = (r - s)x 2 = 0 = r = s. Hence, R im by the first isomorphism theorem (theorem 7.3) = Hereafter, we identify R and im via the map r rx , for r R. x Note that rx = ry , whenever y = 0, so im is independent of x. x y
Andrew Mathas (University of Sydney) Rings, Fields and Galois Theory
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Math3962 -- Lecture 9
Integral domains and fields
Recall that a ring R is an integral domain if it is commutative and it contains no zero divisors. Hence, by Theorem 9.2 and Proposition 9.4 we have the following:
Polynomials over an integral domain
Suppose that R is an integral domain. If f (x) =
i fi x i
deg f (x) =
Corollary 9.5
Every integral domain embeds into a field. Exercise Show that Z[x] is an integral domain.
R[x] then define the degree of f (x) to be -, if f (x) 0, = d, if f (x) = 0 and d = max { i N : fi = 0 }
Lemma 9.7
Suppose that f (x) and g(x) are nonzero polynomials in R[x]. Then f (x)g(x) = 0 and deg f (x)g(x) = deg f (x) + deg g(x). Proof Write f (x) = d fi x i and g(x) = e gi x i , i=0 i=0 where d = deg f (x) and e = deg g(x). = f (x)g(x) = f0 g0 + (f1 g0 + f1 g0 )x + + fd ge x d+e . Now fd = 0 and ge = 0, so fd ge = 0 as R has no zero divisors = deg f (x)g(x) = d + e = deg f (x) + deg g(x) In particular, f (x)g(x) = 0.
Example 9.6
As R = Z[x] is an integral domain it embeds into its field of fractions Z(x) =
f (x) g(x)
: f (x), g(x) Z[x] with g(x) = 0 .
Exercise Show that the cancellation law holds in an integral domain R. That is, if rt = st, for r , s, t R and t = 0, then r = s.
Andrew Mathas (University of Sydney) Rings, Fields and Galois Theory Math3962 -- Lecture 9 5/9
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Andrew Mathas (University of Sydney)
Rings, Fields and Galois Theory
Math3962 -- Lecture 9
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Polynomials over an integral domain.../2
Corollary 9.8
Suppose that R is an integral domain. Then R[x] is an integral domain. Proof Since R is a commutative ring with one the polynomial ring R[x] is a commutative with one (with identity the constant polynomial 1(x) = 1). By the Lemma R[x] has no zero divisors, so R[x] is an integral domain.
Nice rings
Given a ring R we now have several ways of constructing new rings from it: R n = { (r1 , . . . , rn ) : ri R } with componentwise addition and multiplication. Matn (R) = { (rij : rij R } with matrix addition and multiplication. The ring End(R) = { f : R - R : f a ring homomorphism } R/I if I is an ideal of R. If R is commutative: R[x], and R[[x]]. If R is an integral domain: R(x) and the field of fractions of R. Usually, we think of the elements of a ring as being our coefficients: the coefficients of a polynomial, a power series, or a rational function; the entries of a matrix; the coefficients of elements in a module (a module over a field is a vector space see Math3967); the coefficients of elements in an algebra (an algebra `ring over a ring' see Commutati...