Mathematics MA1060 / MA1055-I
2013/2014
Some notes on fields and their basic properties
Let (F, +, ·) be a field. Briefly, the axioms are:
A1: the sum of any two elements in the field also lies in the field
A2: commutativity of addition
A3: associativity of addition
A4: existence of an additive identity, 0
A5: every element in the field has an additive inverse
M1: the product of any two elements in the field also lies in the field
M2: commutativity of multiplication
M3: associativity of multiplication
M4: existence of a multiplicative identity, 1
M5: every element in the field, except 0, has a multiplicative inverse
D: The Distributive Law, multiplication distributes over addition.
Here are some immediate consequences of the basic axioms of a field.
a. −(−x) = x, any x in F . Since −x is the additive inverse of x, it is the number that,
when added to x, gives 0, i.e. x + (−x) = 0. This last equation also says that x is the
number that, when added to −x gives 0, so x is the additive inverse of −x.
b. If x + y = x + z then y = z. Add the additive inverse of x to both sides and use
associativity, A3:
(−x) + (x + y) = (−x) + (x + z) =⇒ (−x) + x + y = (−x) + x + z.
Thus, 0 + y = 0 + z and we can use A4 to conclude that y = z.
This rule allows for cancellation in addition. As a consequence, any x has exactly one
(has a unique) additive inverse. In fact, if x + y = 0 = x + z, then y = z (= −x).
c. −(x + y) = (−x) + (−y), for any x and y in F . We check, using commutativity of
addition A2, that
(−x) + (−y) + (x + y) = (−x) + x + (−y) + y = 0 + 0 = 0.
Hence (−x) + (−y) is the additive inverse of x + y.
d. x.0 = 0 for any x in F . By A4, 0 + 0 = 0. So, by D,
x.0 + 0 = x.0 = x.(0 + 0) = x.0 + x.0.
Cancelling x.0 from each side gives 0 = x.0.
N.B. This explains why 0 cannot have a multiplicative inverse (i.e. there can’t be an x
such that x.0 = 1 unless 0 = 1, in which case (?) the field is trivial).
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e. Denote the multiplicative inverse of x (x 6= 0) by x−1 or 1/x. Then 1/(1/x) = x and
(xy)−1 = x−1 y −1 any x and y in F \ 0. These are the multiplicative versions of a and c
above, with analogous proofs.
f.
If x.y = 0 then either x = 0 or y = 0. If xy = 0 and x 6= 0 then x has a multiplicative
inverse and x−1 (xy) = x−1 .0 = 0, by d above. But x−1 (xy) = (x−1 x)y = 1.y = y (using
associativity of multiplication M3) and we conclude that y = 0.
g. If x.y = x.z and x 6= 0 then y = z. This is the cancellation rule for multiplication,
corresponding to c above. To prove it, multiply both sides by x−1 .
h. (−1)x = −x, any x in F . Notice that
0 = [1 + (−1)].x = (1.x) + ((−1).x) = x + ((−1).x).
This shows that (−1).x acts like the additive inverse of x and so is the additive inverse
of x, i.e. −x.
i.
(−x)y = −(xy) = x(−y) and (−x)(−y) = xy for any x, y in F . By h (twice), (−x)y =
((−1).x)y = (−1)(xy) = −xy. (Alternatively, check that (−x)y acts like the additive
inverse of xy.) By commutativity, x(−y) = −(xy). Finally, by what we’ve just shown,
(−x)(−y) = −(x(−y)) = −(−(xy)) = xy, by a above.
This property is often described as minus by plus is minus, plus by minus is minus,
minus by minus is plus.
Subtraction and Division. Instead of x + (−y) we usually write x − y and refer to this as
subtracting y from x, so that subtraction corresponds to adding the additive inverse. Instead
of x.y −1 we often write x/y or x ÷ y and refer to this as dividing x by y, so that division
corresponds to multiplying by the multiplicative inverse. In this form, the axioms A5 and
M5 become x − x = 0 and x ÷ x = 1, respectively.
Now we add in the axioms for order, less than or <, which are:
O1:
O2:
O3:
O4:
Trichotomy – for any x and y in F , precisely one of x < y, y < x, x = y holds
Transitivity – if x < y and y < z then x < z
If x < y then x + z < y + z for any x, y, z in F
If 0 < x and 0 < y then 0 < xy.
We often write y > x instead of x < y, and write x ≤ y or y ≥ x for ‘x < y or x = y’.
Some consequences of these laws of order are:
j. x > 0 if and only if −x < 0, any x in F . If 0 < x then we can add −x to both sides and
use O3 to conclude that
0 + (−x) < x + (−x) = 0, so that − x < 0.
k. If x < y and z > 0 then xz < yz. Since x < y, it follows that y − x > 0 (add −x to both
sides). Then, since z > 0, O4 gives 0 < z(y − x) = zy − zx. Then, using O3, zx < zy as
required.
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One can similarly show that If x < y and z < 0 then xz > yz. If z < 0 then, by j,
−z > 0. Then k above gives (−z)x < (−z)y which becomes, by i, −(zx) < −(zy). The
result follows by adding zx + zy to both sides.
j.
x2 > 0 for any non-zero x in F . It follows from i that (−x)2 = x2 . Now, for any non-zero
x, either x > 0 or x < 0 (this is O1). If x > 0 then, by 04, x2 = x.x > 0 as well. If x < 0
then, by j, −x > 0 and again by O4, (−x)2 > 0. Once again, x2 > 0. We see that, in
either case, x2 > 0.
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