Algebra by L. E. Sigler (auth.)

By L. E. Sigler (auth.)

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A) + (-a) E S. e E s. Every X E s is +-invertible since - x E S. Both + and · are associative operations on R and therefore certainly on S. Addition is commutative on R and therefore also commutative on S. Thus (S, +, ·, 0) is a ring, a subring of (R, +, ·, ~D Every ring contains as subrings the trivial subring {0} and the entire ring itself. Given an arbitrary subset A of a ring (R, +, ·, 0) we ask the question whether or not there exist subrings of R which contain the given A as a subset. The answer is yes, there always exists at least one subring of R containing A, namely R itself.

In a symmetric manner we prove ( -x)y = -(xy). 0 Theorem. Let (R, +, ·, fJ) be a ring. Then (- x)(- y) = xy for all x, y e R. PRooF. Using the previous theorem twice: ( -x)(- y) = - [x(- y)] = - [- (xy)]. But - (- (xy)) = xy because the inverse of the inverse of any o element is the element itself. The previous theorems demonstrate how some of the frequently performed manipulations of school algebra are valid in rings in general. We also can observe as we develop the theory of rings for what reasons our manipulations of school algebra are valid.

S, t)[(u, v) + (w, x)] + w, v + x) = (s(u + w), t(v + x)) = (su + sw, tv + tx) = (su, tv) + (sw, tx) = (s, t)(u, v) + (s, t)(w, x). = (s, t)(u The right distributive theorem is verified similarly. This completes the verification that (71. , +, ·, (0, 0), (1, 1)) is a unitary commutative ring. The construction in the previous example motivates the following definition of a product ring. Definition. Given rings (R', + ', ·', e') and (R", + ", ·", e") we define the product of the two rings to be ( R' x R", +, ·, (e', e")) in which + and ·are defined by (x', x") + (y', y") = (x' +' y', x" +" y") (x', x") · (y', y") = (x' ·' y', x" ·" y").

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