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Solutions — Ark 1 From the book: Number 8, 9, 10 and 12 on page 11. Number 8 : Show that the set of prime ideals of A has a minimal element with respect to inclusion. Solution: By Zorn’s lemma it sufficies to show that any descending chain of prime ideals contains a prime ideal. Let {pi }i∈I be such a chain, i.e., the set I is linearly ordered and pi ⊆ pj whenever i ≤ j. If the chain is finite, there is a smallest ideal in the chain, which is prime,� so we may assume the chain � to be infinit. We shall see that i pi is prime. Let xy ∈ i pi . Then for a given index i, either x ∈ pi or y ∈ pi . It follows that either x or y must be contained in infinitly many of the pi ’s, and because the chain is descending, the one that is, is contained in the intersection. � √ Number 9 : Let a be a proper ideal in a ring A. Show that a = a is equivalent to a being an intersection of prime ideals. √ Solution: Since the radical a is a, � √ the intersection of all the primes containing clearly a is such an intersection if a = a. Conversly, assume that a = i pi . Then all the pi are among the prime ideals containing a, of course, and hence we have the inclusions � √ a⊆ a⊆ pi . As a = � i pi it follows that a = √ i a. � Number 10 : Let A be a ring with nilradical N. Then the following are equivalent: 1. A has exactly one prime ideal; 2. every element of A is either a unit or nilpotent; 3. A/N is a field. Solution: . The point is that all three are equivalent to N being a maxiamal ideal.Clearly this is equivalent to iii). Now N is the interection of all prime ideals in A, so if A has only one prime ideal, that ideal must be N and, being the only one, it is of course maximal. On the other hand, if N is maximal, it is the only prime ideal, since it is contained in any other. Solutions Ark 1 MAT4200 — autumn 2011 If N is maximal, and hence the only prime ideal, A is a local ring. Elements in A not in N are therefore invertible, and those in N are nilpotent, so elements in A are either invertible or nilpotent. On the other hand, if every non-nilpotent element is invertible, N is a maximal ideal. This shows that ii) is equivalent to N being maximal. � Number 12 : A local ring contains no idempotents other than 0 or 1. Solution: Let e ∈ A be an idempotent, i.e., e2 = e. Then e(e − 1) = 0. Now the ring being local, either e or e − 1 is invertible (both can not lie in the maximal ideal). It follows that either e = 0 or e − 1 = 0. � Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible. (Hint: Use that k[X1 , . . . , Xn ] is UFD.) Solution: In fact, we are going to show that in any ring A being a UFD an element f is irreducible if and only if the principal ideal (f ) is prime. The easy implication — which does not use that A is UFD, and hence is true in general — is that (f ) prime implies f irreducible. To see that, assume that f = xy. Then xy ∈ (f ). Hence, (f ) being prime by assumption, one, of them, say x, is in (f ), and consequently x = af . Substituting back into f = xy we get f = af y, and cancelling f (which we safely can do since A is a domain) we obtain 1 = ay. This shows that y is a unit. To prove the other way around, assume f is irreducible and pick elements x and y with xy ∈ (f ). Then xy = af . As our ring A is a UFD, we may factor x, y and a into irreducibles: x = x1 · · · · · xr , y = y1 · · · · · ys and a = a1 · · · · · at . This gives the two factorisations of xy into irreducibles: xy = x1 · · · · · xr y1 · · · · · ys = a1 · · · · · at f (1) Such factorisations being essentially unique, f has to be a unit times either one of the xi ’s or one of the yj ’s. In the first case x ∈ (f ) and in the second y ∈ (f ). � Oppgave 2. Let k be a field . Show that the ideal (XY − ZW ) ⊂ k[X, Y, Z, W ] is prime. Is it maximal? (Hint: Use the previous exercise and take a look at the degrees of the polynomials involved). Let x, y, z and w cosets of the variables X, Y, Z and W in the quotientring A = k[X, Y, Z, W ]/(XY − ZW ). Show that x is irreducible inA, but that (x)A is not prime. Is k[X, Y, Z, W ]/(XY − ZW ) UFD? —2— Solutions Ark 1 MAT4200 — autumn 2011 Solution: We shall show that (XY − ZW ) is irreducible, and then appeal to the previous exercise. Assume therefore that P Q = (XY − ZW ). We may write P = P0 + P1 + · · · + Pn and Q = Q0 + Q1 + · · · + Qm where the Pi ’s and the Qi ’s are homogenous polynomials of degree i. If Pn and Qm both are not constant, it follows that n + m = 2 since (XY − ZW ) is homogenous of degree 2. Hence either m = 0 or n = 0 and we are through, or n = m = 1 and P and Q are linear forms. Putting Z = 0, one sees that W is not occuring in a nonzero term of either P or Q, putting W = 0 gives that Z does not occure, so either P = X and Q = Y or vise versa. Both cases are absurd. The ideal (XY − ZW ) is not maximal e.g., since (XY − ZW ) ⊂ (X, Y ) which is a proper ideal. Let us see that x is irreducible. Assume x = f g and � pick representatives � F and G in k[X, Y, Z, W ] for f and g respectively. Then F = si=0 Fi and G = ti=1 Gi where Fi and Gi are homogeneus polynomials of degree i. We are allowed to freely change both F and G with elements of the ideal (XY − ZW ), so we may assume that no term Fi or Gi is in (XY − ZW ). We get: � s �� t � � � X = F G + C(XY − ZW ) = Fi Gi + C(XY − ZW ) i=0 = � i<s and j<t i=1 Fi Gj + Fs Gt + C(XY − ZW ). Now if s + t > 1, Fs Gt + Cs+t−2 (XY − ZW ) being the only term of degree s + t in the equation must vanish: Fs Gt + Cs+t−2 (XY − ZW ) = 0. Hence either Fs or Gt has (XY − ZW ) as a factor (the polynomial ring is UFD!) contrary to the hypothesis that no Fi or Gi is in (XY − ZW ). It follows that either F or G is a constant. The ring A is not UFD — in fact in some sense it is the “generic” or minimal example of a non-UDF — since x is irreducible, but not prime. ( We have that zw ∈ (x) but neither z nor w lies in (x).) � √ √ √ Oppgave 3. Let Z[i 5] = {a + ib 5 | a, b ∈ Z} ⊂ C. Show that Z[i 5] is a subring of C and that there is an isomorphism √ Z[X]/(X 2 + 5) � Z[i 5] (2) √ given by X �→ i 5. √ √ Show that 2 · 3 � = (1 + i 5)(1 − i 5) and that the four numbers involved all� are irreducible in Z[i 5]. (Hint: Let N (z) = z z̄ and check that an element w ∈ Z[i 5] is a unit if and only if N (w) = 1.) —3— Solutions Ark 1 Is Z[i MAT4200 — autumn 2011 � 5] a UFD? √ √ √ Show that a = (2, 1+i 5) ⊂ Z[i 5] is a maximal ideal. (Hint: Check that Z[i 5]/(2, 1+ √ i 5) � Z[X]/(2, 1 + x) � F2 � Vis at a2 = (2). Hence (2) = a2 . √ Solution: It is easy to check that Z[i 5] is a subring og the complex field C — it is closed under multiplication and addition. √ √ There is a ring homomorphism Z[X] → Z[i 5] sending P (X) to P (i 5). It is √ obviously surjective. If P (i 5) = 0 it follows from the fundamental theorem of algebra that P (X) = (X 2 + 5)(a0 X n + a1 X n−1 + · · · + an ) where ai ∈ R. Multiplying out, on gets ai+2 + 5ai ∈ Z which gives a√ on i. i ∈ Z by induction √ It is trivial that 2 · 3 = √ (1 + i 5)(1 − i 5) We check that w ∈ Z[i 5] is a unit if and only if N (w) = 1. In that case ww̄=1, so w̄ is an invers for w. If ww� = 1 it follows that N (w)N (w� ) = 1, and since the norm always is a non-negative integer, it follows that N (w) = 1. Now, if wz√ = 2, we get N (w)N (z) = 4 and as a2 + 5b2 = 2 has no solution in Z, no element in Z[i 5] has norm 2. Hence either w or z is of norm 1 and thus a unit. √ In a similar way, a2 + 5b2 = 3 has no integral solution, so no element i Z[i 5] has norm 3. This means that if zw = 3, either z or w is of norm 1 and hence a unit.√ √ As N (1 ± i 5) = 6 it follows in the same way as for 2 and 3 that 1 ± i 5 is irreducible. We have √ just produced a factorisation into irreducibles of an element in two different ways, so Z[i 5] is not UFD. √ The invers image of the ideal a under the map Z[X] → Z[i 5] above is (2, √X + 1, X 2 + 5) = (2, X + 1) since X 2 + 5 = (X + 1)(X − 1) + 3 · 2. Hence Z[i 5] ≈ Z[X]/(2, X + 1, X 2 + 5) =√Z[X]/(2, X + 1)√≈ Z/(2)Z √ = F2 which is a field. 2 2 √We have a = √ (2, 1 + i 5) = (4, 2 + 2i 5, −4 + 2i 5) = (2) since 2 = −4 − (−4 + 2i 5) + (2 + 2i 5). � —4—