I have a little confusion regarding the following fact :
In a ring R if ab = ac then why can't we conclude that b = c ?
Help will be appreciated.
let's find a ring for which this is not TRUE:
let R = Z_{12}, the integers mod 12.
then (2)(7) = 2 (since 14 = 2 (mod 12)), and (2)(1) = 2, but clearly 7 does not equal 1 (mod 12).
perhaps a bit later you will discover that in a commutative ring, the property that ab = ac implies b = c, is equivalent to the property that ab = 0 implies a = 0, or b = 0.
in other words, cancellation in commutative rings requires that we have no "zero-divisors" (zero-divisors can be thought of as "bad elements" which cannot be inverted, just like in the ordinary rational numbers 0 cannot be inverted).
it turns out that in the integers mod n, this separation is CLEAN: if an element is NOT a zero-divisor, it IS invertible:
non-unit = zero-divisor
unit = non-zero-divisor
in general rings, however, we get more kinds of things:
units, zero-divisors, elements which are neither.
it is NOT true that invertiblity is the REASON we cannot "cancel the a". for example, the ring Z (the integers) does not have multiplicative inverses (except for 1 and -1), but we nevertheless DO have cancellation:
for NON-ZERO integers a,b,c:
ab = ac implies b = c (perhaps you can see why we must choose non-zero numbers, and the analogy with zero-divisors, here).
the same is ALSO true of POLYNOMIALS with integer coefficients: if p(x),q(x),and r(x) are not 0-polynomials, then:
p(x)q(x) = p(x)r(x) implies q(x) = r(x). for example:
suppose q(x) = a + bx + cx^{2}, and r(x) = a' + b'x + c'x^{2}, and we know that:
(x - 1)q(x) = (x - 1)r(x).
so (x - 1)(a + bx + cx^{2}) = (x - 1)(a' + b'x + c'x^{2})
ax + bx^{2} + cx^{3} - a - bx - cx^{2} = a'x + b'x^{2} + c'x^{3} - a' - b'x - c'x^{2}
-a + (a - b)x + (b - c)x^{2} + cx^{3} = -a' + (a' - b')x + (b' - c')x^{2} + c'x^{3}
since these are equal, they have equal coefficients, so:
-a = -a'
a - b = a' - b'
b - c = b' - c'
c = c'
since -a = -a', a = (-1)(-a') = (-1)(-a') = a'.
and: b = (b - c) + c = (b' - c') + c = (b' - c') + c' = b'. thus q(x) = r(x) (and no "dividing" was required).
rings are "tricky", we have more algebra at our disposal, but we must remind ourselves that we don't have ALL of the "usual" arithmetic of numbers to use.
Thank you so much. Your explanation was really good ! Cleared a lot of things in my head !
Could you also tell me what are real valued functions of a real variable and real valued functions of a single real variable ? My ring theory book uses them to give examples of rings and subrings, but i don't quite understand what they mean .
a real-valued function of a real variable is a mapping f:R-->R
normally, such a mapping is given by f(x) = formula in x, such as:
f(x) = cos(x) + 3x^{2}.
the set of all such functions (it is a very LARGE set, by the way) can be turned into a ring by defining:
f + g is the function that has the value f(x) + g(x) at x:
(f + g)(x) = f(x) + g(x).
fg is the function that has the value f(x)g(x) at x:
(fg)(x) = f(x)g(x)
for example, if f is given by f(x) = x^{2}, and g is given by g(x) = x+2 then:
(f+g)(x) = x^{2}+x+2 and
(fg)(x) = x[SUP]3+2x^{2}.
the ring axioms are then satisfied as a consequence of the fact that the real numbers themselves form a ring.