MATH 32012 SOLUTIONS SECTION A

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MATH32012 SOLUTIONS SECTION A A1. All rings in this question are understood to be commutative. (a) Give the definition of a noetherian ring. A noetherian ring is a ring where every ideal is finit ... ely generated. (b) Give the definition of an irreducible element of a ring. An element r of a ring R is irreducible, if r 6= 0, r is not invertible and r = ab implies that a or b is invertible. (c) Does there exist a noetherian ring which does not contain any irreducible elements? Justify your Yes, for example a field K. It is noetherian because its only ideals are <0> and K = <1>. Every element of K is either zero or invertible, hence not irreducible. Now let R[X, Y ] denote the ring of polynomials in two variables X, Y with real coefficients. (d) State a reason why every irreducible element of R[X, Y ] is a prime element of R[X, Y ]. Because R[X, Y ] is a unique factorisation domain. (e) Give an example of a prime element of R[X, Y ] which is not irreducible. Justify your example. 0, which is not irreducible by definition, and is prime because 0|fg =⇒ 0 = fg =⇒ 0 = f or 0 = g (R[X, Y ] is a domain) =⇒ 0|f or 0|g. (f) Is X4 + Y 4 an irreducible element of R[X, Y ]? Justify your No, X4 + Y 4 = (X2 + Y 2)2 - 2X2Y 2 = (X2 + Y 2 - √2 XY )(X2 + Y 2 + √2 XY ), a product of two non-invertible elements. (g) Write the polynomial X4 - X4Y 4 + 2X2Y 2 - 3XY 3 - √2 Y 5 ∈ R[X, Y ] in standard form with respect to each of the monomial orderings: (1) ≺Lex with X ≻ Y ; (2) ≺Deglex with X ≻ Y . (h) Reduce f = Y 4 + XY 3 with respect to the set F = {f1 = Y 2 - XY 2, f2 = Y 2 - Y 3} using the monomial ordering ≺Deglex with X ≻ Y . Express f in the form h1f1 + h2f2 + r for some h1, h2 ∈ R[X, Y ], where r is the remainder that you found. With respect to the chosen monomial ordering, the leading term of f1 is -XY 2 and the leading term of f2 is -Y 3. At each step, we will underline the term in the polynomial 2 of 6 P.T.O. MATH32012 SOLUTIONS which is divisible by the leading term of f1 or of f2. One has The remainder is r = 2Y 2 and moreover f = -Y f1 - (Y + 2)f2 + r. [NB: the remainder, r, is uniquely determined as {f1, f2} is a Gr¨obner basis.] In parts (i)–(l), you DO NOT need to justify your answers. (i) Give an example of a maximal ideal of R[X, Y ]. For example, <X, Y >. (j) Let I be the ideal of R[X, Y ] generated by all polynomials of the form XmY n where m, n ∈ Z, m > 0 and n ≥ (m - 3)2. Give an example of a finite set T ⊂ R[X, Y ] such that I = <T >. For example, T = {XY 4, X2Y, X3}. To arrive at this answer, recall that a monomial ideal is generated by its minimal monomials, and the minimal monomials form a finite set by Dickson’s Lemma. One can work out the minimal monomials of I with the aid of a diagram (black dots represent monomials in I): (k) Identify the variety V(I) where I is the ideal defined in part (j). V(I) = V({XY 4, X2Y, X3}) is the same as V({X}), i.e., the line {(0, t) : t ∈ R} ⊂ R2. (l) Let C = {(cos t, sin t) ∈ R2 : t ∈ R, 0 ≤ t < 2π}. Identify the ideal I(C) of R[X, Y ]. C is the circle of radius 1 centred at the origin, and I(C) = <X2 + Y 2 - 1>. MATH32012 SOLUTIONS SECTION B B2. Let R = K[X1, X2, . . . , Xn] where K is a field, and let ≺ be a monomial ordering for R. (a) Give the definition of a Gr¨obner basis with respect to ≺ of an ideal of R. A Gr¨obner basis of an ideal I is a finite subset {f1, f2, . . . , fn} of I such that the leading monomials of the fi with respect to ≺ generate the ideal LT≺(I) of leading terms of I. (b) What is a reduced Gr¨obner basis? A reduced Gr¨obner basis is a Gr¨obner basis {g1, . . . , gn} such that all the gi are monic polynomials and for any i 6= j, the polynomial gi is reduced with respect to gj. (c) Let f1 = XY - X2, f2 = X - Y 2 in Q[X, Y ], and let I be the ideal of Q[X, Y ] generated by f1 and f2. With respect to the monomial ordering ≺Lex with X ≻ Y , (i) compute a Gr¨obner basis for I; (ii) find a reduced Gr¨obner basis for I. Buchberger algorithm: start with G = {XY -X2, X-Y 2}. We will double-underline leading monomials with respect to ≺Lex. Compute the S-polynomial S(XY - X2, X - Y 2) = XY - X2 + X(X - Y 2) = XY - XY 2. Reduce it with respect to G (when reducing, we single-underline a monomial divisible by a leading monomial of one of the polynomials in Compute the S-polynomials and reduce them with respect to G: All the S-polynomials reduce to 0, so G = {XY - X2, X - Y 2, Y 3 - Y 4} is a Gr¨obner basis. A short way to find a reduced Gr¨obner basis is as follows: the Gr¨obner basis found above shows that the ideal of leading terms is <X2, X, Y 4>, and that X2 can be deleted from the generating set as it is divisible by X. (Alternatively, XY - X2 can be explicitly reduced to 0 with respect to the other two polynomials.) Therefore, {X -Y 2, Y 3 -Y 4} is a Gr¨obner basis. No reductions are possible, so, multiplying by constants to make all the polynomials monic, we obtain the reduced Gr¨obner basis G′ = {X - Y 2, Y 4 - Y 3}. [NB: this is the only possible answer as the reduced Gr¨obner basis with respect to the given monomial ordering is unique.] 4 of 6 P.T.O. https://www.coursehero.com/file/9237825/2011-model-solutions/ MATH32012 SOLUTIONS (d) Prove that the ideal I, defined in part (c), does not contain any monomials in Q[X, Y ]. Any element of I is of the form f = h1f1 + h2f2 where h1, h2 ∈ Q[X, Y ]. Note that the zero set V(I) = V({f1, f2}) ∋ (1, 1) as f1(1, 1) = f2(1, 1) = 0. But no monomial XmY n vanishes at the point X = Y = 1, thus no monomial belongs to I. B3. (a) What is meant by saying that a commutative ring R is a principal ideal domain? R is a domain and every ideal of R can be generated by one element. (b) Denote by S the set of all rational numbers of the form p q where p is an integer and q is an odd integer. Show that S, with the standard operations of addition and multiplication of numbers, is a ring. It is enough to check that S is a subring of Q. Closure under addition: if p, q, r, s ∈ Z, q, s are odd, p q + r s = ps + qr qs ∈ S because qs is odd. Closure under multiplication: p q r s = pr qs ∈ S because qs is odd. Furthermore, S contains 0 = 0 1 and 1 = 1 1 (the denominator, 1, is odd) and is closed under negation: -p q = -p q ∈ S where q is odd. (c) Find the set U(S) of units of the ring S defined in part (b). p q ∈ S is invertible in S if and only if q p ∈ S, if and only if p is odd. Thus U(S) = {p q : p, q are odd integers }. (d) Find the set of all irreducible elements of the ring S. The set of irreducible elements of S is {2p q : p, q are odd integers }. Indeed, each element of this set is not zero, and is not invertible by (c). If 2p q = r s t u where p, q, s, u are odd, then 2psu = qrt. The left-hand side is not divisible by 4, so r, t cannot both be even; therefore, one of r, t is odd, so one of r s , t u is a unit. This proves that 2p q is irreducible. Any element not in this set is either a unit or of the form 4p q where p, q ∈ Z, q is odd; as 4p q = 2 • 2 p q is a product of two non-units, this is not irreducible. (e) Prove that the ring S is a principal ideal domain. First of all, S is a domain because it is a subring of a domain Q. Every non-zero element of S can be written as 2k p q where p, q are odd integers and k ∈ Z, k ≥ 0. Thus, every element is associated to 2k, k ∈ Z, k ≥ 0. Let I be an ideal of the ring S. If I = {0}, then I is principal. Otherwise, with every non-zero element 2k p q , the ideal I contains its associate 2k. Let 2k0 be the lowest power of 2 contained in I. Then <2k0> ⊆ I, and for every 2k p q ∈ I one 5 of 6 P.T.O. / MATH32012 SOLUTIONS has 2k p q = (2k-k0 p q )2k0 ∈ <2k0>, so I ⊆ <2k0>. Thus, I = <2k0> is principal. Remark. In fact, this ring is euclidean with euclidean norm N(2k p q ) = k where p, q are odd integers. B4. Let R be a commutative ring. (a) Define the notion of a nilpotent element of R. An element a ∈ R is nilpotent if there is n ∈ N such that an = 0. (b) Let Nil(R) denote the set of all nilpotent elements of R. Prove that Nil(R) is an ideal of R. (Note for the 2012 exam: it is enough to say that Nil(R) = √<0> therefore it is an ideal.) (c) (i) Define the radical, √I, of an ideal I of R. √I = {a ∈ R | ∃n ∈ N : an ∈ I}. (ii) Show that pI + Nil(R) = √I. As I + Nil(R) ⊇ I, one has pI + Nil(R) ⊇ √I. On the other hand, I ⊆ √I, and Nil(R) = p{0} ⊆ √I as {0} ⊆ I. Therefore, I + Nil(R) ⊆ √I and pI + Nil(R) ⊆ p√I = √I. (d) Prove that if x is a nilpotent element of R, then 1 - x is a unit of R. Let n be such that xn = 0. Then 1 = 1 - xn = (1 - x)(1 + x + x2 + . . . + xn-1) which shows that 1 - x is invertible. (e) Give an example of a finite ring R such that Nil(R) consists of 2011 elements. Justify your example. (Hint: 2011 is a prime number.) For example, R = Z20112. This ring consists of residues of {0, 1, . . . , 20112 - 1} modulo 20112. The invertible elements of this ring are residues of a such that a is co-prime with 20112, i.e., a is not divisible by 2011. Invertible elements are not nilpotent. The noninvertible elements of Z20112 are the residues of 2011k for 0 ≤ k < 2011. That is, there are 2011 non-invertible elements; and each of them squares to zero, as (2011k)2 is divisible by 20112. Thus, there are 2011 nilpotent elements in Z20112. END OF EXAMINATION PAPER 6 of 6 Powered by TCPDF (www.tcpdf.org) [Show More]

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