Bourbaki

Exercise 24, § 1

Algebra, VI, p. 33

Let A be a commutative ring and I the set of all ideals of A. Then a(b+c)=ab+aca(b + c) = ab + ac for all a,b,ca, b, c in I, by I, p. 107; in other words I, with the order relation a⊇ba \supseteq b and the law of composition (a,b)↦ab(a, b) \mapsto ab, is a semi-lattice ordered monoid (VI, p. 31, Ex. 13), the sup of two elements a, b of I being a∩ba \cap b, and their inf a+ba + b.

a) Let K be a field and A=K[X,Y]A = K[X, Y] the ring of polynomials in two indeterminates over K. Consider the principal ideals a=(X),b=(Y),c=(X+Y)a = (X), b = (Y), c = (X + Y) in A. Show that

(a∩b)+c≠(a+c)∩(b+c), (a \cap b) + c \neq (a + c) \cap (b + c),
(a+b)∩b≠(a∩c)+(b∩c), (a + b) \cap b \neq (a \cap c) + (b \cap c),
(a∩b)(a+b)≠(a(a+b))∩(b(a+b)) (a \cap b)(a + b) \neq (a(a + b)) \cap (b(a + b))

b) In the ring A, 1 is a gcd of X and Y, but (X)+(Y)≠A(X) + (Y) \neq A.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.