# Algebra. Rings, modules and categories by Carl Faith By Carl Faith

VI of Oregon lectures in 1962, Bass gave simplified proofs of a few "Morita Theorems", incorporating rules of Chase and Schanuel. one of many Morita theorems characterizes while there's an equivalence of different types mod-A R::! mod-B for 2 jewelry A and B. Morita's resolution organizes rules so successfully that the classical Wedderburn-Artin theorem is an easy final result, and in addition, a similarity type [AJ within the Brauer workforce Br(k) of Azumaya algebras over a commutative ring okay comprises all algebras B such that the corresponding different types mod-A and mod-B together with k-linear morphisms are identical via a k-linear functor. (For fields, Br(k) involves similarity periods of straightforward primary algebras, and for arbitrary commutative okay, this is often subsumed less than the Azumaya 1 and Auslander-Goldman [60J Brauer crew. ) a number of different situations of a marriage of ring thought and classification (albeit a shot­ gun wedding!) are inside the textual content. in addition, in. my try and additional simplify proofs, significantly to get rid of the necessity for tensor items in Bass's exposition, I exposed a vein of rules and new theorems mendacity wholely inside of ring idea. This constitutes a lot of bankruptcy four -the Morita theorem is Theorem four. 29-and the root for it's a corre­ spondence theorem for projective modules (Theorem four. 7) instructed by way of the Morita context. As a spinoff, this offers origin for a slightly whole thought of easy Noetherian rings-but extra approximately this within the advent.

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Extra info for Algebra. Rings, modules and categories

Example text

So it follows that α = x 2 = 32 = 9. According to this we transform x + x1 as follows A=x+ 1 x 1 8 x 1 8 2 8 10 = + + x≥2 · + x = + ·3= . 8 Let a, b > 0 be real numbers such that a + b ≤ 1. Find the minimum value of the expression A = ab + 1 . e. ab = 1. and equality occurs if and only if ab = ab √ But then we have a + b ≥ 2 ab = 2, contradicting a + b ≤ 1. 1 1 4 4 , then we have x = ab ≥ (a+b) If we take x = ab 2 ≥ 12 = 4. Thus we may consider an equivalent problem of the given problem: Find the minimum of the function A = x + x1 , with x ≥ 4.

So let us assume that A, B > 0. e. 5). Since QM ≥ GM we have |a1 b1 + a2 b2 + · · · + an bn | ≤ |a1 b1 | + |a2 b2 | + · · · + |an bn | a 2 + b12 a22 + b22 a 2 + bn2 ≤ 1 + + ··· + n 2 2 2 (a12 + a22 + · · · + an2 ) + (b12 + b22 + · · · + bn2 ) = 1, = 2 as required. Equality occurs if and only if ab11 = ab22 = · · · = abnn . ) Proof 2. Consider the quadratic trinomial n n n (ai x − bi )2 = i=1 (ai2 x 2 − 2ai bi x + bi2 ) = x 2 i=1 n ai2 − 2x i=1 n ai bi + i=1 bi2 . e. 2 n 4 ai bi n −4 i=1 i=1 2 n ⇔ ai bi i=1 as required.

Xn ) = f (x1 , x2 , . . , xn ) − g(x1 , x2 , . . , xn ) is homogenous. In other words, a given inequality is homogenous if all its summands have equal degree. 2 The inequality x 2 + y 2 + 2xy ≥ z2 + yz is homogenous, since all monomials have degree 2. The inequality a 2 b + b2 a ≤ a 3 + b3 is also homogenous, but the inequality a 5 + 5 b + 1 ≥ 5ab(1 − ab) is not homogenous. In the case of a homogenous inequality, without loss of generality we may assume additional conditions, which can reduce the given inequality to a much simpler form.