Let x be a nilpotent element of a ring A and u be a unit of A. Then u + x is a unit.

Transcription

1 1 Rings and Ideals 1.1. Let x be a nilpotent element of a ring A and u be a unit of A. Then u + x is a unit. Proof. Since x is nilpotent, it is in every prime ideal, so that x + u cannot be in any prime ideal of R, hence must be a unit. To find an explicit formula for the inverse of u + x, note that u + x = u(1 + u 1 x). If y = u 1 x, then (1 + y) 1 is given by 1.2. ( 1) n y n = n=0 m ( 1) n y n n=0 Where m is the least positive integer with y m = 0. Let R be a ring and f R[x] and f = a 0 + a a n x n. Then (a) If f is a zero-divisor, there is an a R (0) with af = 0. (b) f is nilpotent if and only each a j is nilpotent (c) f is a unit if and only a 0 is a unit and a 1,..., a n are nilpotent Proof. (a) Let h be a polynomial of minimal degree with hf = 0 and h = b 0 + b 1 x b m x m, so that b m 0. Then a n b m = 0, hence the degree of a n h is strictly less than that of h and a n hf = 0, thus we must have a n h = 0. We wish to show that a n s h = 0 for all s with 0 s n. Thus a n s b m = 0, hence b m f = 0 and b m is nonzero. The case s = 0 has already been established. Suppose that 0 < s n and for all t < s, we have a n t h = 0. The coefficient of x n s+m in hf is 0 (since hf = 0) and is given by m+n s j=0 a j b m+n s j If j < n s, then b m+n s j = 0, thus we need only concern ourselves with the sum But this sum is equal to m+n s j=n s a j b m+n s j m a n+i s b m i i=0 1

2 By hypothesis, a n s+i = 0 for all i > 0. Thus all that is left is the element a n s b m. Thus a n s b m = 0, as needed. (b) = If f is nilpotent, then a 0 is nilpotent, hence f a 0 is nilpotent, and thus a 1 is nilpotent and hence, a 1 x is nilpotent. By induction, we find that each a j is = If each a j is nilpotent, then a j x j is nilpotent. Thus f is the sum of nilpotent elements, hence nilpotent. (c) = If f is a unit, then a 0 is clearly a unit. If S is ring that is an integral domain, then the units in S[x] consist of the units of S. Hence, for any prime ideal p R, R/p is an integral domain. In (R/p)[x] = R[x]/p[x], the image of f is a unit, hence a j p for j > 0. Thus for j > 0, a j lies in every prime ideal of R, hence is nilpotent. = For j > 0, a j x j is nilpotent, hence a 1 x +... a n x n is nilpotent and a 0 is a unit, hence f = a 0 + a 1 x +... a n x n is a unit by (1.1) In the ring A[x] the Jacobson radical is equal to the nilradical. Proof. Let f = a i x i A[x] be in the Jacobson radical of A[x]. Then for every i=0 g A[x], 1 fg is a unit. In particular, taking g = x, by (1.2) a 0, a 1,..., a n are nilpotent, hence f is nilpotent, since it is a sum of nilpotent elements Let A[[x]] denote the formal power series ring over A. Let f = (a) f is a unit in A[[x]] if and only if a 0 is a unit in A a n x n. Then (b) If f is nilpotent, then a n is nilpotent for all n 0. Is the converse true? (c) f belongs to the Jacobson radical of A[[x]] if and only if a 0 belongs to the Jacobson radical of A (d) The contraction of a maximal ideal m of A[[x]] is a maximal ideal of A and m is generated by m c and x. (e) Every prime ideal in A is the contraction of a prime ideal in A[[x]] n=0 2

3 Proof. (a) One direction is clear. If a 0 is a unit, we construct a g = b n x n such that fg = 1 by induction. For n = 0, let b 0 = a 1 0. For n > 0, suppose that we have constructed b i for all i < n. For g to satisfy fg = 1, the coefficients of f and g must satisfy the equation n=0 However, a n j b j = j=0 ( n 1 a n j b j = 0 j=0 ) a n j b j + a 0 b n. Thus, j=0 b n = a 1 0 ( n 1 ) a n j b j j=0 (b) If f A[[x]] is nilpotent, then a 0 is nilpotent, hence f a 0 is nilpotent. Thus a 1 is nilpotent, hence a 1 x nilpotent, so that f (a 0 + a 1 x) is nilpotent. Continue. The converse is false. For p Z prime consider the ring And power series A = Z/p n Z n=1 f = a j x j j=0 Where a 0 = a 1 = 0 and a j is the sequence (b nj ) defined by b nj = 0 for n j and b jj = p+p j Z. Then j is the least positive integer m such that a m j = 0. Then or r s, a r a s = 0. Thus for any m f m = a m j x mj j=0 Thus there is no m such that f m = 0, since there is no m N such that a m j = 0 for all j 0. (c) f Jrad(A[[x]]) if and only if 1 fg is a unit for all g A[[x]] if and only if 1 a 0 b is a unit for every b A. (d) Since 1 xh is a unit in A[[x]] for every h A[[x]], we have xa[[x]] Jrad(A[[x]]). If there was an ideal I of A such that m c I A, choose y I such that y / m c. Then ya[[x]] + m = A[[x]]. Choose f A[[x]] and g m such that yf + g = 1. Where 3

4 f = g = a n x n n=0 b n x n n=0 Then ya 0 + b 0 = 1. Since xa[[x]] Jrad(A[[x]]), we have g = b n x n m. Thus b 0 m, hence b 0 m c. I = A and m c is maximal. Thus for any g m, the constant term of g is in m c and x m, hence m = (x, m c ). (e) For p Spec(A), p is the contraction of the prime ideal p[[x]] of A[[x]]. Here, p[[x]] denotes the set of all f A[[x]] with coefficients in p. n= Let A be a ring such that every ideal not contained in the nilradical contains a nonzero idempotent element. Then the Jacobson radical and the nilradical are equal. Proof. If the nilradical and the Jacobson radical were not equal, there would be a nonzero idempotent e Jrad(A). However, e(1 e) = and e Jrad(A), hence 1 e is a unit, but then e = 0, a contradiction Let A be a ring in which every element satisfies x n = x for some n > 1 depending on x. Then every prime ideal is maximal. Proof. Suppose that x / p and that n > 1 such that x n = x. Then x(x n 1 1) p, thus x n 1 1 p. Since n > 1, x is a unit in A/p. Hence p is maximal Let A be a nonzero ring. Then A has a minimal prime ideal. Proof. Since A 0, Spec(A) is nonempty. We may partially order Spec(A) by reverse inclusion. That is for p, q Spec(A), p q if and only if q p. By Zorn s lemma, it suffices to show that each chain (p λ ) λ Λ has an upper bound. Since any maximal element of Spec(A), ordered by reverse inclusion, is clearly a minimal prime ideal of A. Let (p λ ) λ Λ be a chain and let p = p λ. Suppose that a / p and b / p, thus there are µ, ν Λ such that a / p µ and b / p ν. Since (p λ ) is a chain, either p ν p µ or p µ p ν. Assume the former, then a / p ν, hence ab / p ν and p is prime. Since p λ p for every λ Λ, p is an upper bound for the chain (p λ ). 4

5 1.8. Let I be a proper ideal of a ring A. Then I = r(i) if and only if I is an intersection of prime ideals. Proof. = The radical of I is the intersection of all prime ideals that contain it. = Write I = p j, where p j Spec(A). Then x n I for some n > 0, then x n is in every p j, hence x is in every p j. Thus I = r(i) Let A be a ring and Nil(A) be its nilradical. Then the following are equivalent: (a) A has exactly one prime ideal (b) every element of A is either a unit or nilpotent (c) A/Nil(A) is a field Proof. (a) = (b) Let p be the unique prime ideal of A. Then Jrad(A) = Nil(A) = p and if x is not a unit, x p, hence is nilpotent. (b) = (c) Every nonzero element of A/Nil(A) is invertible, hence A/Nil(A) is a field. (c) = (a) The nilradical is the intersection of all prime ideals of A, hence any two prime ideals of A must be equal A ring A is said to be Boolean if x 2 = x for all x A. Let A be Boolean, then (a) 2x = 0 for all x A (b) Every prime ideal p is maximal and A/p is a field with two elements. (c) Every finitely generated ideal is principal. Proof. (a) Since (x + x) 2 = (x + x), we have The statement follows. x 2 + x 2 + x 2 + x 2 = x + x + x + x = x + x (b) For x / p, we have x 2 x p, since x 2 = x. Thus x 1 p, hence A/p is a finite integral domain (consisting of two elements) and thus a field. 5

6 (c) By induction on the number of generators, it suffices to prove the case in which an ideal is generated by two elements. Suppose I = (x, y)a. Then x = x(x + y xy) and y = y(x + y xy), hence I = (x + y xy)a Let A be a nonzero ring and Σ the set of ideals in which every element is a zero divisor. Then Σ has maximal elements and every such maximal element is prime. Proof. Σ is nonempty (0 Σ) and Zorn s lemma applies to easy to obtain a maximal element p. Suppose that x / p and y / p. The ideals p + xa and p + ya contain elements that are not zero divisors, since they strictly contain p. Say a p + xa and b p + ya are not zero divisors, then ab is not a zero divisor, hence xy / p, since p consists entirely of zero divisors Let A be a ring and X be the set of all prime ideals of A. For each subset E of A let V (E) denote the set of all prime ideals which contain E. Then (a) If I is the ideal generated by E then V (E) = V (I) = V (r(i)) (b) V (0) = X and V (1) = (c) If E α is a family of subsets, V ( E α ) = V (E α ) (d) V (IJ) = V (I J) = V (I) V (J), for ideals I and J of A Proof. (a) E I, so that V (I) V (E). If p is a prime ideal containing E, then p contains all R-linear combinations of E, hence I p. Thus V (I) = V (E). Since I r(i), V (r(i)) V (I). If p is a prime ideal of A with I p, then r(i) r(p) = p. Thus V (I) = V (r(i)). (b) 0 is in every prime ideal and 1 is not in any prime ideal (c) V ( E α ) = V ( E α ) and for a prime ideal p of A, E α p if and only if E α p for every α. (d) Since IJ I J, so that V (I J) V (IJ). If p V (IJ), then I p or J p. In either case, I J p. Thus V (I J) = V (IJ). If p V (I) V (J), then p contains I or J, hence I J and conversely, if I J p, then p contains I or J. Thus V (I) V (J) = V (I J) For each f A let X f denote the complement of V (f) in X = Spec(A). The sets X f are open and form a basis of open sets for the Zariski topology and 6

7 (a) X f X g = X fg (b) X f = if and only if f is nilpotent (c) X f = X if and only if f is a unit (d) X f = X g if and only if r(f) = r(g). (e) X is compact (f) Each X f is compact (g) An open subset of X is compact if and only if it is a finite union of sets X f. Proof. For any p X, p A, so there is an f A p. Suppose now that q Xis such that q X f X g. Then f / q and g / q, hence fg / q and q X fg. Clearly, X fg X f X g. (a) For p X, p X fg if and only if fg / p if and only if f / p and g / p if and only if p X f X g. (b) f is nilpotent if and only if V (f) = X if and only if X f = (c) f is a unit if and only if V (f) = if and only if X f = X (d) X f = X g if and only if V (f) = V (g) if and only if every prime ideal containing f also contains g and vice versa, hence if and only if r(f) = r(g). (e) Suppose that X = j J X fj, where f j A. Let I be the ideal generated by all the f j. If I were a proper ideal, there would be a maximal ideal m such that I m, but then m contains all the f j, contradicting that m f j for some j J. Thus there are finitely many f j, say f 1,..., f n such that there is an equation of the form a i f i = 1 i=1 Where the a i A. Given p X, at least one of the f i cannot be in p, since p is n proper. Thus p and X is compact. i=1 X fi (f) Suppose X f X gj, where the g j A. Then V (g j ) V (f). Let I be j J j J the ideal generated by all the g j in A. Then for p X with I p, we have that f p, hence f r(i). Thus there are finitely many g j and an n > 0 such that there is an equation of the form f n = a i g i Thus if f / p, there is at least one g i such that p / g i. Hence, X f i=1 7 n i=1 X gi

8 (g) A finite union of compact sets is always compact. Conversely, if U is a compact open set of X, then U is a union of sets of the form X f. Since U is compact, there can be only many such X f Let X = Spec(A) for a ring A and let p X. Then (a) The set {p} is closed in X if and only if p is maximal (b) Cl({p}) = V (p) (c) q Cl({p}) if and only if p q. (d) If q and p are two distinct prime ideals of X, then either there is a neighborhood of p which does not contain q or there is a neighborhood of q which does not contain p. That is, X is a T 0 -space. Proof. (a) Suppose {p} = V (I) for some ideal I A. If m is a maximal ideal such that I m, then m = p. Conversely, if p is maximal, {p} = V (p). (b) If q Cl({p}), then q is in every closed set containing {p}, hence is in V (p). Conversely, if q V (p), then q V (I) for for every closed set V (I) such that {p} V (I). (c) This follows directly from (b). (d) We may assume there is an f p q. Then q X f and p / X f A topological space X is said to be irreducible if X is nonempty and every pair of nonempty open sets in X intersect, or equivalently if every nonempty open set is dense in X. Then X = Spec(A) is irreducible if and only if the nilradical is a prime ideal. Proof. X is irreducible if and only if X f X g = X fg for all f, g A, since every open set of X is a union of sets of the form X h, h A. However, X f X g = X fg if and only if f / Nil(A) and g / Nil(A) implies that fg / Nil(A). The last statement is equivalent to the nilradical being a prime ideal of A Let X be a topological space. (a) If Y is an irreducible subspace of X, then the closure Cl(Y ) of Y in X is irreducible. (b) Every irreducible subspace of X is contained in a maximal irreducible subspace 8

9 (c) The maximal irreducible spaces of X are closed and cover X. They are called the irreducible components of X. What are the irreducible components of a Hausdorff space? (d) If A is a ring and X = Spec(A), then the irreducible components of X are the closed sets V (p), where p is a minimal prime ideal of A. Proof. (a) Cl(Y ) is nonempty, since Y is nonempty by definition of irreducible. Let Cl(Y ) A and Cl(Y ) B be two nonempty open subsets of Cl(Y ), where A and B are open sets in X. Since Cl(Y ) A is nonempty, Y A is nonempty as well and similarly, Y B is not empty as well. We have Y A B (Cl(Y ) A) (Cl(Y ) B) and thus (Cl(Y ) A) (Cl(Y ) B) is not empty. Cl(Y ) is irreducible. (b) Let Σ denote the set of all irreducible subspaces of X that contain Y ordered by inclusion. Then Σ is nonempty, since Y Σ. Let (W j ) j J be a chain in Σ and let W = j J W j. W is nonempty since each W j is nonempty. Let A, B be nonempty open subsets of X, so that W A and W B are nonempty. We have, W A = (W j A) and W B = (W j B). Thus there is an i and j such that W i A and W j B. Since (W j ) is chain, we may assume that W i W j, hence W j A is nonempty. Since W j is irreducible, (W j A) (W j B) is nonempty and (W j A) (W j B) (W A) (W B), hence W is irreducible. Thus Zorn s lemma yields a maximal irreducible subspace of X containing Y. Thus the maximal irreducible subspaces of X are the singleton sets. (c) If Y is maximal irreducible subspace of X, then Cl(Y ) is irreducible and Y Cl(Y ), hence Y = Cl(Y ) and Y is closed. For x X, the set {x} is clearly irreducible and there is a maximal irreducible subspace Y x such that {x} Y x. Thus, Y = x X Y x. If X is Hausdorff and x X, then {x} Y for some maximal irreducible subspace Y of X. If we had y Y such that y x, then there are disjoint open neighborhoods U and V such that x U and y V. Then U Y and V Y, hence Y (U V ), since Y is irreducible. However, this contradicts U V =. (d) Note that a topological space is irreducible if and only if it cannot be written as the union of two proper closed subsets. Let I A be an ideal and suppose V (I) were irreducible. Since V (I) = V (r(i)), we may assume that I = r(i). Suppose now f, g A such that fg I. Then V (I) = (V (f) V (I)) (V (g) V (I)). Since V (I) is irreducible, we may assume that V (I) = V (I) V (f), hence V (I) V (f), thus f r(i) = I. Thus I is prime. Conversely, suppose that I is prime ideal, if we have ideals J, J of A such that V (I) = V (J) V (J ). Then V (J) V (J ) = V (JJ ), hence JJ r(i) = I, then either J I or J I, since I is prime. Thus V (I) is irreducible. Suppose now Y is a maximal irreducible subset of X. Then Y is closed, hence Y = V (p) for some ideal p of A. By the above, p is a prime ideal. If there were a 9

10 q X such that q p, then V (p) V (q). Since V (p) is a maximal irreducible subset of X and V (q) is irreducible, V (p) = V (q), hence p = q. Thus p is a minimal prime ideal of A Let ϕ : A B be a ring homomorphism. Let X = Spec(A) and Y = Spec(B). If q Y, then ϕ 1 (q) = q c is a prime ideal of A, that is, a point of X. Hence ϕ induces a mapping ϕ : Y X. Then (a) If f A, then ϕ 1 (X f ) = Y ϕ(f), thus ϕ is continuous (b) If I is an ideal of A, then ϕ 1 (V (I)) = V (I e )) (c) If J is an ideal of B, then Cl(ϕ (V (J))) = V (J c ). (d) If ϕ is surjective, then ϕ is a homeomorphism of Y onto the closed subset V (Ker(ϕ)) of X. (e) ϕ (Y ) is dense in X if and only if ker(ϕ) Nil(A). (f) Let ψ : B C be another ring homomorphism. Then (ψ ϕ) = ϕ ψ (g) Let A be an integral domain with one nonzero prime ideal p and let K be the quotient field of A. Let B = (A/p) K. Define ϕ : A B by ϕ(x) = (x + p, x). Then ϕ is a bijection but not a homeomorphism. Proof. (a) For p Y, p ϕ 1 (X f ) if and only if f / p c if and only if ϕ(f) / p. The sets X f form a basis for the topology on X, hence ϕ is continuous. (b) For p Y, if p I e, then p c I ec I. Conversely, if p c I, then p p ce I e. (c) Since ϕ (V (J)) V (J c ), we have Cl(ϕ (V (J))) V (J c ). Suppose now p V (J c ) and choose f A such that p X f. Then f / r(j c ) = r(j) c. Thus there is q Y such that q J and f / q c. Hence, X f ϕ (V (J)), thus p Cl(ϕ (V (J))). (d) Let I = ker(ϕ). Then A/I = B, so that ϕ provides a bijection between V (I) and Y. Thus it remains to show that ϕ 1 : V (I) Y is continuous. For g B, let f A such that ϕ(f) = g. Then (ϕ 1 ) 1 (Y g ) = ϕ (Y g ) = ϕ (Y ϕ(f) ) = X f V (I). Thus ϕ 1 is continuous and ϕ is a homeomorphism. (e)= If there were if an f ker(ϕ) Nil(A), there p X such that f / p. However, X = Cl(ϕ (Y )), hence X f ϕ (Y ). Thus there is a q Y such that f / q c, but ϕ(f) = 0, a contradiction. Hence, ker(ϕ) Nil(A). = Let p X and choose p X f. Then ϕ(f) / Nil(B), since ϕ is injective and f / Nil(A). Thus there is a q Y such that ϕ(f) / q. Hence, X f ϕ (Y ). 10

11 (f) Let p Spec(C). Then (ψ ϕ) (p) = (ψ ϕ) 1 (p) = ϕ 1 (ψ 1 (p)) = ϕ (ψ (p)). (g) By (1.18), Spec(B) = {0 K, (A/p) 0}. We have ϕ (0 K) = {x A : (x + p, x) 0 K} = p ϕ ((A/p) 0) = {x A : (x + p, x) (A/p) 0} = 0 Thus ϕ is a bijection. Then (ϕ 1 ) 1 ({(A/p) 0}) = ϕ ({(A/p) 0}) = {0} However, {0} is not closed, since 0 is not a maximal ideal in A. Thus ϕ 1 is not continuous, hence ϕ is not a homeomorphism If A is a ring and A = A 1 A 2, then every ideal of A is of the form I 1 I 2, where I j is an ideal of A j. Hence, Spec(A) = {A 1 p : p Spec(A 2 )} {q A 2 : q Spec(A 1 )}. Proof. Let I A be an ideal. Then I = J 1 J 2, where J s A s are subsets and clearly 0 J s, since I is an ideal. Let I 1 = {a A 1 : (a, 0) I} I 2 = {a A 2 : (0, a) I} Then I s is an ideal of A s and J s = I s. The statement follows. For the second statement, given P Spec(A), P = I 1 I 2 for ideals I j of A j and A/P = A 1 /I 2 A 2 /I 2. However, A/P is an integral domain and this is true if and only if one of the I j = A j and the other is a prime ideal of the respective ring Let A be a ring. Then the following are equivalent: (a) X = Spec(A) is disconnected (b) A = A 1 A 2 where neither A 1 are both nonzero. (c) A contains a nontrivial idempotent. Proof. (a)= (b) Since X is disconnected, there are nonempty closed proper subsets of X, say V (I) and V (J) such that X = V (I) V (J) and V (I) V (J) =. Since V (I) V (J) = V (IJ) = X, IJ Nil(A). If there were a prime ideal p such that I + J p, then p V (I) V (J), a contradiction. Thus I + J = A. Now let x + y I + J be such that x + y = 1. Since xy IJ, there is an n > 0 such that 11

12 (xy) n = 0. The ideals x n A + y n A are comaximal, so there are a, b A such that ax n + by n = 1. Then (1 ax n ) 2 = (1 ax n )by n = by n = 1 ax n Then 1 ax n is neither 0 nor 1, since both x and y are not units. Lastly, given a nontrivial idempotent e of A, it is easy to see that A = ea (1 e)a and hence, A is isomorphic (as a ring) to (A/eA) (A/(1 e)a). (b) = (c) Clear. (c) = (a) Let f be a nontrivial idempotent of A. Any nilpotent idempotent must be 0, so that we may assume that f / Nil(A). Thus there is a prime ideal p such that f / p, by (1.13), X f is a proper subset of X. Since 1 f is also idempotent and f is nontrivial, 1 f is neither a unit nor nilpotent, hence X 1 f is a nontrivial proper subset of X. Clearly, X f X 1 f = and there is no q X such that q / X f and q / X 1 f. Hence, X = X f X 1 f and X is disconnected Let A be a Boolean ring and let X = Spec(A). Then (a) For each f A, X f is both open and closed. (b) Let f 1,..., f n A. Then X f1 X fn = X f for some f A. (c) The sets X f are the only sets which are both open and closed. (d) X is a Hausdorff space Proof. (a) For f A, f(f 1) p for any prime ideal p A and f / p if and only if 1 f p. Hence, X f = V (1 f). (b) By induction, we reduce to the case n = 2. Let f, g A. Then by (a), X f X f = V (1 g) V (1 f) = V ((1 f)(1 g)) = X 1 (1 f)(1 g) (c) Let Y X be both open and closed. From (1.13), X is compact and the sets X f form a basis for the topology on X, hence Y is a finite union of sets of the form X f. By (b), Y = X g for some g A. (d)let p and q be distinct prime ideals of A. We may assume there is an f p q. Then p V (f) = X 1 f and q X f and X f X 1 f =. X is Hausdorff Suppose that I Nil(A) is an ideal. If a A is such that a + I is a unit in A/I, then a is a unit in A. 12

13 Proof. There is a b A such that ab = 1 + y, where y Nil(A). By (1.1), ab is a unit in A, hence a is a unit in A Let A and B be rings and f : A B a surjective ring homomorphism. (a) f(jrad(a)) Jrad(B). Give an example to show inclusion may be strict. (b) If A is semilocal, then f (Jrad(A)) = Jrad(B). Proof. (a) Let x Jrad(A) and let b B. Then b = f(a), hence 1 bf(x) = f(1 ax). Since x Jrad(A), 1 ax is a unit, hence 1 bx is a unit. Thus, f(x) Jrad(B). Let A = Z and B = Z/4Z and f : A B be the natural map. Jrad(A) = 0 and Jrad(B) = 2B, hence inclusion is strict in this case. Clearly, (b)let I = ker(f), so that A/I = B and the maximal ideals of B are in a bijective correspondence with the maximal ideals of A that contain I. Let m 1,..., m n be the maximal ideals of A and they can be numbered so that m 1,..., m s contain I and m s+1,..., m n do not contain I. Then Jrad(A) = m 1 m n and f(jrad(a)) = f(m 1 m n ) = f(m 1 ) f(m s ) = Jrad(B) Let A be a ring in which every prime ideal is principal, then every ideal in A is principal. Proof. By Zorn s Lemma, it suffices to show that an ideal which is maximal with respect to being non-principal is a prime ideal of A. To this end, let I A be an ideal which is maximal with respect to being non-principal and a, b A such that a / I and b / I. Let I be the ideal generated by I and a and I the ideal generated by I and b. By the maximality of I, both I and I are principal. Write I = xa. Suppose now that ab I and let J = (I : I ). Then I I J, so we can write J = za, hence I J = (xz)a I. Let y I. Since I I, hence y = y x, but then r J. Write y = y z, so that x I J and hence, I J = I, contradicting that I is not principal. I is a prime ideal Let A be a ring and x be an element of A such that x is in every nonzero ideal of A. Let I = ann A (x). Then I is a maximal ideal of A and if y A I, then y is a nonzero divisor. 13

14 Proof. Suppose that a / I. Then ax 0, hence x (ax)a. Write x = abx, so that 1 ab I. Thus A/I is a field, hence I is a maximal ideal. Let y A I and a A (0) such that ay = 0. Then x aa, hence we can write x = ba for some b A. Then xy = b(ay) = 0, a contradiction, hence y is not a zero-divisor Let A be an infinite integral domain with a finite number of units. Then A has an infinite number of maximal ideals. Proof. Let x Jrad(A). Then given an infinite set of distinct nonzero elements of A, say {a j }, 1 a j x is a unit for each j. Since A has only finitely many units, 1 a j x = 1 a i x for some i j. Thus we must have x = 0. If A has only finitely many maximal ideals, then Jrad(A) = m 1 m n = 0 and A = A/m 1... A/m n. However, A is an integral domain, so n = 1, but then A is an infinite field, contradicting that A has only finitely many units Let e 1,..., e n be idempotents of a ring A. Let s m (e 1,..., e n ) = e j1... e jm be the m-th symmetric polynomial in the e i. Then f = 1 j 1 <j 2...<j m n ( 1) m+1 s m (e 1,..., e n ) is m=1 idempotent and if a m A are such that a = a m e m, then fa = a. Proof. By induction. The case n = 1 is clear. We have the following identities: m=1, And for 1 < m < n, s 1 (e 1,..., e n ) = s 1 (e 1,..., e n 1 ) + e n s n (e 1,..., e n ) = s n 1 (e 1,..., e n 1 )e n s m (e 1,..., e n ) = s m (e 1,..., e n 1 ) + s m 1 (e 1,..., e n 1 )e n Thus ( 1) m+1 s m (e 1,..., e n ) can be written as m=1 ) n 1 n 1 ( 1) m+1 s m (e 1,..., e n 1 ) + e n (1 ( 1) m+1 s m (e 1,..., e n 1 ) m=1 m=1 14

15 n 1 Then x = ( 1) m+1 s m (e 1,..., e n 1 ) is idempotent by the induction hypothesis m=1 and e n is idempotent. Thus, f = x+e n (1 x) and f 2 = x 2 +2x(1 x)e n +e 2 n(1 x) 2 = x + e n (1 x) = f. Then Let b = n 1 m=1 e m a m, so that a = b + e n a n and xb = b by the induction hypothesis. fa = (x + e n (1 x))(b + e n a n ) = xb + xe n a n + be n (1 x) + e 2 na n (1 x) = b + xe n a n + e n a n (1 x) = b + a n e n = a Let A be a ring in which every proper ideal is prime. Then A is a field. Proof. A is an integral domain. Let x be a nonzero element of A. If x is not a unit, then x 2 A is a prime ideal. However, then x x 2 A, but this contradicts that x is not a unit, since A is an integral domain. Hence, x is a unit and A is a field. 15