1.4 A little algebra
A ring is a set \(R\) equipped with two maps
usually written as \((x,y)\mapsto x + y\), and
usually written as \((x,y)\mapsto x\cdot y\) or as \((x,y)\mapsto xy\), along with elements \(0,1\in R\) satisfying the following properties:
For all \(x,y,z\in R\), \((x+y) + z = x + (y+z)\).
For all \(x,y\in R\), \(x+y = y+x\).
For all \(x\in R\), \(x+0 = x\).
For all \(x\in R\), there exists \(y\in R\) such that \(x + y = 0\).
For all \(x,y,z\in R\), \((xy)z = x(yz)\).
For all \(x\in R\), \(x1 = 1x = x\).
For all \(x, y, z\in R\), \(x(y+z) = xy + xz\) and \((x+y)z = xz + yz\).
The element \(y\) in d is unique. In other words, if \(R\) is a ring and \(x + y = x + y' = 0\), then \(y = y'\).
We have \(y' = 0 + y' = y' + 0 = y' + (x + y) = (y' + x) + y = (x + y') + y = 0 + y = y\).
The element \(y\) in d is called the negative of \(x\) and written as \(-x\).
The axioms in the definition of a ring are useful because there are so many examples of rings.
The set \(\mathbb {Z}\) of integers forms a ring with the usual operations of addition and multiplication along with the usual \(0\) and \(1\). So do the sets \(\mathbb {Q}\) of rational numbers and the set \(\mathbb {R}\) of real numbers.
A ring \(R\) is said to be commutative if, for all \(x,y\in R\), we have \(xy = yx\). We say that \(R\) is noncommutative if it is not commutative.
The rings \(\mathbb {Z}\), \(\mathbb {Q}\) and \(\mathbb {R}\) are all commutative. On the other hand, let \(M_2(\mathbb {R})\) denote the set of all \(2\times 2\)-matrices
with \(a\), \(b\), \(c\) and \(d\) in \(\mathbb {R}\). Let \(+\) and \(\cdot \) be the usual matrix addition and matrix multiplication. Let \(0\in M_2(\mathbb {R})\) denote the \(0\) matrix and let \(1\) by the identity matrix
Then \(M_2(\mathbb {R})\) is a noncommutative ring.
Suppose \(R\) is a ring and \(a\in R\). An element \(b\in R\) is said to be a multiplicative inverse of \(a\) if \(ab = ba = 1\). An element \(a\in R\) is said to be invertible if it has a multiplicative inverse. We write \(R^{\times }\) for the set of all invertible elements of \(R\).
Suppose \(R\) is a ring and \(x\) and \(y\) are in \(R\).
If \(x\in R^{\times }\), then there exists a unique multiplicative inverse \(x^{-1}\) of \(x\) in \(R\).
If \(x\), \(y\in R^{\times }\), then \(xy\in R^{\times }\) and \((xy)^{-1} = y^{-1}x^{-1}\).
(a) Suppose \(y\) and \(z\) are both multiplicative inverses of \(x\). Then \(y = y1 = y(xz) = (yx)z = 1z = z\).
(b) Suppose \(x\), \(y\in R^{\times }\). Then \((xy)(y^{-1}x^{-1}) = xy(y^{-1}x^{-1})= x(yy^{-1})x^{-1} = x1x^{-1} = xx^{-1} = 1\). And, similarly, \((y^{-1}x^{-1})(xy) = 1\) as well.