Elementary Abstract Algebra- Examples and Applications, 2019a

840 CHAPTER 20 INTRODUCTION TO RINGS AND FIELDS
In the following proposition, we show that we may construct the additive
inverse of any ring element by multiplying the element by the additive inverse
of 1.
Proposition 20.1.24. Let R be a ring, let −1 be the additive inverse of 1,
and let −x denote the additive inverse of x ∈ R. Then−x =(−1) · x.
□
Exercise 20.1.25. Prove Proposition 20.1.24.
♦
Since there are many rules that ring operations obey, we can simplify
algebraic expressions in rings in much the same way as we do in basic algebra.
Exercise 20.1.26. Given A, B, C ∈ R where R is a commutative ring, we
can show that (B +(−C)) · (A · (B + C)) = A · (B · B +(−C) · C)). Give
the reasons for the following steps in the simplification:
(B +(−C)) · (A · (B + C)) = (A · (B + C)) · (B +(−C)) Comm. prop.
= A · ((B + C) · (B +(−C))) Assoc. prop.
= A · (((B + C) · B)+((B + C) · (−C))) < 1 >
= A · ((B · B + C · B)+(B · (−C)+C · (−C))) < 2 >
= A · ((B · B + B · C)+((−C) · (B)+(−C) · C)) < 3 >
= A · (B · B +(B · C +(−C) · (B)+(−C) · C)) < 4 >
= A · (B · B +(B · C + B · (−C)) + (−C) · C)) < 5 >
= A · (B · B +(B · (C +(−C)) + (−C) · C)) < 6 >
= A · (B · B +(B · (0) + (−C) · C)) < 7 >
= A · (B · B +(0+(−C) · C)) < 8 >
= A · (B · B +(−C) · C)) < 9 >
♦
20.2 Subrings
Earlier we mentioned that two important topics studied in Abstract Algebra
are groups and rings. Just like groups have subgroups, rings have subrings.
Definition 20.2.1.
subring.
A ring that is a subset of another ring is called a
△