{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "\n", "\n", "
\n", "\n", "\n", "\n", "\n", "\n", "Abstract Algebra: An Interactive Approach, 2e\n", "
\n", "\n", "\n", "©2015 This notebook is provided with the textbook, "Abstract Algebra: An Interactive Approach, 2nd Ed." by William Paulsen. Users of this notebook are encouraged to buy the textbook.\n", "
\n", "\n", "\n", "\n", "\n",
"Chapter 9
\n",
"Introduction to Rings\n",
"
Initialization: This cell MUST be evaluated first:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "%display latex\n", "try:\n", " load('absalgtext.sage')\n", "except IOError:\n", " load('/media/sf_sage/absalgtext.sage')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "x · y = 0
\r\n", "where x and y are both nonzero? If so, how many examples can you find?i·j = k, j·k = i, and k·i = j.
\r\n", "This suggests that we can also add multiples of these elements together like vectors, forming such elements asProperty | \r\n", "ℤ | \r\n", "Even integers | \r\n", "ℚ | \r\n", "Reals | \r\n", "Z6 | \r\n", "Quaternions | \r\n", "
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Closed under addition | \r\n", "x | \r\n", "x | \r\n", "x | \r\n", "x | \r\n", "x | \r\n", "x | \r\n", "
Closed under multiplication | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
(a + b) + c = a + (b + c) | \r\n",
" \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
(a·b)·c = a·(b·c) | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
Additive identity (0) | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
Multiplicative identity (1) | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
Additive inverses exist | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
Multiplicative inverses exist except for 0 | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
a + b = b + a | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
a·b = b·a | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
a·b = 0 only if a or b = 0 | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
(a + b)·c = a·c + b·c | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
a·(b + c) = a·b + a·c | \r\n", "\r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " | \r\n", " |
(a + b)·c = (a·c) + (b·c)
\r\n", "and \r\n", "a·(b + c) = (a·b) + (a·c)
\r\n", "hold for all a, b, and c in the ring.x·e = e·x = x
\r\n", "for all elements x in the ring is called a unity ring or ring with identity. The element e is called the unity or \r\n", "multiplicative identity of the ring, to distinguish it from that additive identity 0.e = e · e′ = e′.
\r\n", "\r\n", "x·y = y·x = e,
\r\n", "we say that x has a multiplicative inverse, or is invertible.0·x = 0, not x.
\r\n", "Likewise, for a non-trival ring, 0 cannot have a multiplicative inverse, for otherwise\r\n", "0·(0-1) = 0,
\r\n", "which we just proved was not the multiplicative identity.n x = (n − 1) x + x.
\r\n", "We also define (−n) x to be −(n x) for n a positive integer. Finally, we define 0 x = 0.(m + n) x = m x + n x and (m n) x = m (n x)
\r\n", "for any element x and any integers n and m.(n x)·y = n (x·y) = x·(n y).
\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "\r\n", "We can now use this notation within Sage to generate a finite ring. For example, suppose that we wish to define a ring whose additive group is isomorphic to \r\n", "Z15* = {1, 2, 4, 7, 8, 11, 13, 14}.
\r\n", "Two elements which generate this group are a = 2 and b = 14. Since \r\n", "24 ≡ 1 (mod 15) and 142 ≡ 1 (mod 15),
\r\n", "a is of order 4 and b is of order 2. But in the new notation, we would write\r\n", "4 a = 0 and 2 b = 0.
\r\n", "since 0 is the additive identity for the ring.n1 x1 + n2 x2 + n3 x3\r\n", "+ ··· + nk xk = 0
\r\n", "for integers n1, n2, n3, … nk\r\n", "is if \r\n", "n1 x1 = n2 x2 = n3 x3\r\n", "= ··· = nk xk = 0.
\r\n", "n1 x1 + n2 x2 + \r\n", "n3 x3 + ··· + nk xk
\r\n", "where each ni is non-negative and less then the order of xi, forms a distinct element. Also, every element of G could be \r\n", "put in that form. Thus, the product of the orders of all the elements of B equals the order of the group.2 a = a + a = b·b + b·b = (b + \r\n", "b)·b = (2 b)·b = 0·b = 0.
\r\n", "\r\n", "An example of a ring definition that does not produce such a contradiction comes from defining a2 = a, b2 = b, and \r\n", "a·b = b·a = 0. All other products in the ring can be determined from these using the distributive law. For example,\r\n", "(2 a + b)·(a + b) = 2 a2 + b·a + \r\n", "2 a·b + b2 = 2 a + 0 + 0 + b = 2 a + b.
\r\n", "\r\n", "[[a·a, a·b], [b·a, b·b]].
\r\n", "To define the ring described in the above paragraph, we can usea·b = a·c,
\r\n", "then b = c. Likewise, if\r\n", "b·a = c·a
\r\n", "for a nonzero and not a zero divisor, then b = c. This is called the cancellation law for multiplication.\r\n", "Type I: The Trivial Ring\r\n", "
\r\n", "\r\n", "\r\n", "Type II: Fields\r\n", "
\r\n", "\r\n", "\r\n", "Type III: Skew Fields\r\n", "
\r\n", "\r\n", "\r\n", "Type IV: Commutative unity rings without zero divisors, but which are not fields\r\n", "
\r\n", "\r\n", "\r\n", "Type V: Non-commutative unity rings without zero divisors, but which are not skew fields\r\n", "
\r\n", "\r\n", "a + b i + c j + d k
\r\n", "where a, b, c, and d are all integers. Certainly the product and sum of two integer quaternions would yield integer quaternions, \r\n", "and so these would form a ring. However, 2 has no inverse in the integer quaternions, so this is not a skew field.\r\n", "\r\n", "\r\n", "Type VI: Commutative rings without unity and without zero divisors\r\n", "
\r\n", "\r\n", "\r\n", "Type VII: Non-commutative rings without unity and without zero divisors\r\n", "
\r\n", "\r\n", "a + b i + c j + d k
\r\n", "where a, b, c, and d are all even integers. It is easy to check that the sum and product of two even quaternions yield even \r\n", "quaternions. However, this ring does not have a unity element.\r\n", "\r\n", "\r\n", "Type VIII: Commutative unity rings with zero divisors\r\n", "
\r\n", "\r\n", "\r\n", "Type IX: Non-commutative unity rings with zero divisors.\r\n", "
\r\n", "\r\n", "\r\n", "Type X: Commutative rings without unity and with zero divisors\r\n", "
\r\n", "\r\n", "\r\n", "Type XI: Non-commutative rings without unity and with zero divisors\r\n", "
\r\n", "\r\n", "+ | \r\n", "0 | \r\n", "a | \r\n", "b | \r\n", "c | \r\n", "
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0 | \r\n", "0 | \r\n", "a | \r\n", "b | \r\n", "c | \r\n", "
a | \r\n", "a | \r\n", "0 | \r\n", "c | \r\n", "b | \r\n", "
b | \r\n", "b | \r\n", "c | \r\n", "0 | \r\n", "a | \r\n", "
c | \r\n", "c | \r\n", "b | \r\n", "a | \r\n", "0 | \r\n", "
· | \r\n", "0 | \r\n", "a | \r\n", "b | \r\n", "c | \r\n", "
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0 | \r\n", "0 | \r\n", "0 | \r\n", "0 | \r\n", "0 | \r\n", "
a | \r\n", "0 | \r\n", "a | \r\n", "a | \r\n", "0 | \r\n", "
b | \r\n", "0 | \r\n", "b | \r\n", "b | \r\n", "0 | \r\n", "
c | \r\n", "0 | \r\n", "c | \r\n", "c | \r\n", "0 | \r\n", "
x-1 = | \r\n", "a | \r\n", "+ | \r\n", "− b | \r\n", "i + | \r\n", "− c | \r\n", "j + | \r\n", "− d | \r\n", "k | \r\n", "
a² + b² + c² + d² | \r\n", "a² + b² + c² + d² | \r\n", "a² + b² + c² + d² | \r\n", "a² + b² + c² + d² | \r\n", "
y = (x-1·x)·y = x-1·(x·y) = \r\n", "x-1·0 = 0.
\r\n", "Thus, if both x≠0 and y≠0, then x·y≠0.(0·x + 0·x) =(0 + 0)·x = 0·x,
\r\n", "so\r\n", "(0·x + 0·x) + (−(0·x)) = 0·x + (−(0·x)) = 0.
\r\n", "Hence\r\n", "0·x + (0·x + (−(0·x))) = 0,
\r\n", "so\r\n", "0·x + 0 = 0·x = 0.
(x·0 + x·0) = x·(0 + 0) = x·0,
\r\n", "so\r\n", "(x·0 + x·0) + (−(x·0)) = x·0 + (−(x·0)) = 0.
\r\n", "Hence\r\n", "x·0 + (x·0 + (−(x·0))) = 0,
\r\n", "so\r\n", "x·0 + 0 = x·0= 0.
x = x·e = x·(y·z) = \r\n", "(x·y)·z = 0·z = 0,
\r\n", "which contradicts the fact that x is nonzero. Thus, a division ring has no zero divisors.((n − 1) x)·y + x·y = (n − 1) (x·y) \r\n", "+ x·y = x·((n − 1) y) + x·y.
\r\n", "Hence, by the distributive law,\r\n", "((n − 1) x + x)·y = ((n − 1) + 1) (x·y) = \r\n", "x·((n − 1) y + y),
\r\n", "and so\r\n", "(n x)·y = n (x·y) = x·(n y).
\r\n", "Hence, the statement is true for all positive integers.(n x)·y + ((−n) x)·y = \r\n", "(n x + (−n) x)·y = ((n − n) x)·y = 0·y = 0.
\r\n", "n (x·y) + (−n) (x·y) = \r\n", "(n − n) (x·y) = 0 (x·y) = 0.
\r\n", "x·(n y) + x·((−n) y) = \r\n", "x·(n y + (−n) y) = x·(( − n) y) = x·0 = 0.
\r\n", "Thus, ((−n) x)·y, (−n) (x·y), and x·((−n) y) \r\n", "are the additive inverses of (n x)·y, n (x·y) and x·(n y), respectively. \r\n", "But since these latter three are equal for positive n, we have\r\n", "((−n) x)·y = (−n) (x·y) = \r\n", "x·((−n) y).
\r\n", "\r\n", "Hence the lemma is proven for all integers n.e = m x + n y
\r\n", "for some integers m and n. Since e commutes with both x and y, we have\r\n", "m x·x + n y·x = (m x + n y)·x = \r\n", "e·x = x·e = m x·x + n x·y,
\r\n", "so n y·x = n x·y.m x·y + n y·y = (m x + n y)·y = \r\n", "e·y = y·e = m y·x + n y·y,
\r\n", "so m x·y = m y·x.u m + v n = gcd(m, n).
\r\n", "If we let c denote the greatest common divisor of m and n, then\r\n", "c(x·y − y·x) = \r\n", "(u m + v n)(x·y − y·x) = \r\n", "u (m x·y − m y·x) + v (n x·y − n y·x) = 0.
\r\n", "x·y − y·x = \r\n", "e·(x·y − y·x) = \r\n", "(a·c·x + b·c·y)·(x·y − \r\n", "y·x) = (a·x + b·y)·(c·(x·y − \r\n", "y·x)) = (a·x + b·y)·0 = 0.
\r\n", "So x·y = y·x, and the ring is commutative.y = y·e = y·(x·z) = \r\n", "(y·x)·z = e·z = z,
\r\n", "which is a contradiction.(x·y)·(y-1·x-1) = \r\n", "x·x-1 = e,
\r\n", "we see that x·y is invertible. The associative law comes from the associative multiplication of the \r\n", "ring. So the set of invertible elements forms a group.(a e)·(b e) = (a b) (e·e) = (a b) e = n e = 0.
\r\n", "\r\n", "But since a and b are both less than n, (a e) and (b e) are both nonzero. \r\n", "Hence, these would both be zero divisors in Zn.(a e)·(b e) = (a b) (e·e) = (a b) e = 0.
\r\n", "This would imply that (a b) is a multiple of n. But since n is prime, we would have to conclude that either \r\n", "a or b is a multiple of n. But this contradicts the fact that both (a e) and (b e) are nonzero. Thus, if n is prime, \r\n", "there are no zero divisors in Zn.0 = a·b − a·c = a·(b − c).
\r\n", "But since a is not a zero-divisor and is nonzero, we must have that b − c = 0. Hence b = c.0 = b·a − c·a = (b − c)·a,
\r\n", "and since a is nonzero and not a zero divisor, b − c = 0, and so b = c.{b1, b2, b3, b4, …}
\r\n", "Since R is finite, two elements of this sequence must be equal, say bm = bn for m < n.\r\n", "Using the law of cancellation, we have bm−1 = bn−1. Continuing this way, we eventually get\r\n", "b = bn−m+1
\r\n", "(It is tempting to use Lemma 9.4 one more time to get e = bn−m, but unfortunately we have yet to prove that R \r\n", "has a unity.)x·ba = x·b,
\r\n", "and since b is nonzero and not a zero divisor, we can use the law of cancellation to get \r\n", "x·ba−1 = x.
\r\n", "Likewise, since ba·x = b·x, we have that ba−1·x = x. \r\n", "Hence, there is a unity element in R, namely ba−1.b·ba−2 = ba−1 = e, and \r\n", "ba−2·b = ba−1 = e.
\r\n", "So ba−2 is the multiplicative inverse of b.a·b = a, b·a = b, \r\n", "a·c = c, and c·a = a.
\r\n", " \r\n", "(Hint: Using the associtivity law, determine what a2, b2, and c2 must be. Then show that \r\n", "c·b must commute with a. Use trial and error to determine b·c.a2 = a + c, a·b = b + c, \r\n", "b·a = b, and c·b = c.
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