Algebra - Ring and Field Theory
1.
Any element in R is of the form I + (ax + b), where I = (x2 + 1), the ideal generated by x2 + 1 and a, b are complex numbers. Hence {I + x, I + 1} is a basis of R over C and hence dimC R = 2, so 1 is false.
Since C is a field, C[x] is a PID. Prime ideals of C[x] containing (x2+1) are (p(x)), where p(x) is irreducible divisor of x2+1 (since (a) ⊆ (b) iff b | a).
Since x + i and x - i are the only prime factors of x2 + 1, C[x] has only two prime ideals containing (x2 + 1). Since there is a one-one correspondence between prime ideals containing (x2+1) and prime ideals of C[x]/(x2 +1),
2 is true.
Since (x2 + 1) is not a prime ideal in C[x], R is not an integral domain,
hence 3 is false.
Since any maximal ideal in R is prime and (x + i) / (x2 + 1) and (x - i) / (x2+1) are the only prime ideals of R, 4 is false. Ans. 2.
2.
.
By Eisenstein Criterion (p = 3), f(x) is irreducible over Q, hence 1 is false.
f(m) = m7 - 105m + 12 = m(m6 - 105) + 12
If m is even then f(m) = m(m6 - 105) + 12 is even.
If m is odd then m6 is odd and m6 - 105 is even, hence f(m) = m(m6 - 105) + 12 is even. Hence f(m) is even for all integer m and 2 is false.
If f(m) is a prime then f(m) = 2, since the only even prime is 2.
Hence m7 - 105m + 12 = 2, i.e m7 - 105m + 10 = 0, hence (x - m) is a factor of
x7 - 105x + 10, which is irreducible over Q, by Eisenstein Criterion.
Hence 3 is false and 4 is true. Ans. 4.
Aliter for 2 :
If 2 is true, m7 - 105m + 12 = 105, i.e m7 - 105m - 93 = 0, hence (x - m) is a factor of x7 - 105x - 93, which is irreducible over Q, by Eisenstein Criterion. Hence 2 is false.
3.
The ring Z4 has only finitely many ideals, but it is not a field.
Any infinite field has only two ideals.
Z[√3] is a subring with identity of the real field R, but since
4 = 2.2 = (1-√3)(1+√3), Z[√3] is not a UFD and hence not a PID.
But R is a PID.
If a is a nonzero element of R then <an+1> ⊆ <an>, for all n. If R is an integral domain which has only finitely many ideals then <an+1> = <an> for some n. Hence an = an+1r, for some r in R and n. Since an is nonzero, by the cancellation, 1 = ar, hence a has the multiplicative inverse in R. Ans. 4.
4.
If f(x)g(x) = 0 then either f(x) = 0 or g(x) = 0. Hence 1 is true.
Suppose I(x) is contained in some larger ideal J and g ∈ J is not in I(x).
Then a = g(x) ≠ 0 and h = (1/a)g ∈ J (since the constant function 1/a is continuous). And (h − 1)(x) = h(x) - 1(x) = (1/a)g(x) - 1 = 1 - 1 = 0,
hence (h − 1) ∈ I(x) ⊆ J. Consequently 1 = h + [1 − h] ∈ J. Thus J = C([0, 1]). Hence I(x) is a maximal ideal.
But there are maximal ideals M ≠ I(x), for any x.
For example, if K = {f ∈ C((0,1)) : f(1/n) = 0 for all but finitely many n} then
K is an ideal and hence K ⊆ M, for some maximal ideal M ⊆ C((0,1)).
But M ≠ I(x), for any x ∈ (0, 1).
Note that for any compact set X, in the ring of all continuous functions from
X → R, all the maximal ideals are of the form
Mγ = {f / f(γ) = 0}, for some γ ∈ K.
Let f(t) = 0, if 0 < t ≤ 1/2 ; t - ½ , if ½ ≤ t < 1 and
g(t) = ½ - t, if 0 < t ≤ 1/2 ; 0, if ½ ≤ t < 1.
Then fg = 0 but neither f = 0 nor g = 0. Hence 4 is false. Ans. 1, 2.
5.
6.
7.
8.
9.
Since R[X] is a UFD, R is a UFD. Since R[X] / I = {I + a / a ϵ R} ≅ R and R is an integral domain, 1 is true. If I is maximal then R[X] / I ≅ R is a field, hence R is a PID and hence R[X] is a PID.
If R[X] is a Euclidean domain then it is a PID. If R[X] is a PID then R is a field, hence I is maximal, also R[X] is a Euclidean domain. Ans. 1, 2, 3, 4.
10.
1 is a property of irreducible polynomials. 2 is obviously true.
The polynomial f(x) = x2 - 3 is irreducible over Z, but it is reducible over R and f(x) = x2 = xx of f(x) modulo 3 is reducible in F3[x]. 3, 4 are false. Ans. 1, 2.
11.
12.
By the property of finite fields, 1 is true.
For a finite field F, the multiplicative group (F, ×) is cyclic but the additive group (F, +) need not be cyclic.
Order of any finite field F is pm, where p is a prime and the characteristic of F is p. Hence px = 0, for all nonzero elements x in F. Since p is a prime, every nonzero element of F has additive order p and hence the order of the additive cyclic group generated by them is p.
So, if m > 1 then (F, +) is not cyclic.
Hence 2 and 3 are false.
Since any two finite fields of the same order are isomorphic, 4 is true. Ans. 1, 4.
13.
A0 = {1}. F27* is a cyclic group. For 𝝰 in F27*, |A𝝰| = 26 if and only if O(𝝰) = 26. The number of 𝝰 s.t. O(𝝰) = 26 is φ(26) = φ(2)φ(13) = 1.12 = 12.
Hence 1 is true.
A1 = {1, 1 + 1, 1 + 1 + 1 = 0}, since 27 = 33, Char F27 = 3. Hence 3 is false.
Lt 𝝰 ≠ 0, 1, and n b the order of 𝝰 .
Then 𝝰n = 1. .e. (1 - 𝝰)(1 + 𝝰 + 𝝰2 + . . . + 𝝰n-1) = 0,
hence 1 + 𝝰 + 𝝰2 + . . . + 𝝰n-1 = 0 and 2 is true.
Since 1 is in A𝝰, for all 𝝰, and A0 = {1}, 4 is true. Ans. 1, 2, 4.
14.
Since the characteristic of Fp2 is p, {0, 1, 2, . . . , p - 1} is a subfield of order p of Fp2. Let S be a subring with identity of order p of Fp2. If a is any nonzero, non identity element in S, since S is finite an = 1 for some n > 1, hence aan--1 = 1 and hence a is a unit. Hence S is a subfield of order p of Fp2.
Since there is only one field of order p, there is only one subring with identity.
(infact, Fpn has a unique subring with unity of order pm where m divides n, since there is a unique subfield of order pm for every m.) Ans. 2.
15.
Since the group Fp × Fp is not cyclic, the order of any non-identity element in Fp × Fp is p. Any non-identity element generates a subgroup of order p.
If H is a subgroup of Fp × Fp of order p and a is in Fp × Fp - H is non identity then K = <a> is a different subgroup of order p. Since there are p2 - 1 and p - 1 non-identity elements in Fp × Fp and in a subgroup of order p respectively, there are (p2 - 1) / (p - 1) = p + 1 subgroups of order p in Fp × Fp .
Since p ≥ 5, 1 is true.
The subgroup <(1, 1)> = {(1, 1), (2, 2), . . . , (p-1, p-1), (0, 0)} is not of the form H × K, where H, K are subgroups of Fp, and it is not an ideal of Fp × Fp , hence 2, 3 are false.
The non zero element (1, 0) has no multiplicative inverse in Fp × Fp , 4 is false. Ans. 1.
16.
17.
Let a = 𝛂𝞯. Then a satisfies the polynomial x5 - 2, which is irreducible over Q. Hence the minimal polynomial of a over Q is x5 - 2, So,
K = Q(a) = {r1 + r2a + r3a2 + r4a3 + r5a4 / ri are in Q} is an extension of degree 5. The roots of x5 - 2 are 𝛂, 𝛂𝞯, 𝛂𝞯2, 𝛂𝞯3, 𝛂𝞯4. Among them only a = 𝛂𝞯 is in K.
If 𝛔 is an automorphism of K then 𝛔(a) should be in K and a root of x5 - 2. Hence 𝛔(a) = a. 4 is true.
Let E = Q(𝛂, 𝞯) be the splitting field of x5 - 2 over Q. Then E is a finite extension of Q containing K. Since K is a subset of E, for all field automorphism 𝛔 of K, 𝛔(K) is contained in E. 3 is true.
Since a=𝛂 𝞯 = 𝛂(1/𝞯) is not in K, for the conjugate field automorphism
𝛔 : z → z of C, 𝛔(K) ≠ K, 2 is true.
Define an automorphism 𝛎 on E by 𝛎(𝛂) = 𝛂 and 𝛎(𝞯) = 𝛂𝞯2.
Then 𝛎 is non identity automorphism of E and 𝛎(𝛂𝞯) = 𝛂2𝞯2, hence 𝛎(K) = K. Since any automorphism of any subfield of the complex field C can be extended to an automorphism of C, the automorphism 𝛎 on E can be extended to an automorphism 𝛔 of C. And 𝛔 is non identity, 𝛔(K) = 𝛎(K) = K.
1 is true. Ans. 1, 2, 3, 4.
18.
Any group of order p2 is abelian and isomorphic to Zp2 or Zp × Zp,
hence 1, 2 are true.
Zp2 , Zp × Zp and Fp2 the finite fields of order p2 are rings of order p2,
3 is false. Since any finite integral domain is a field and there is only one field of order p2, 4 is true. Ans. 1, 2, 4.
19.
If a is a root of p(x) with multiplicity > 1, then a is a root of p′(x) also.
If p = 2 then p(x) = x4 + x + 0 = x4 + x and p′(x) = 4x3 + 1 = 0 + 1 = 1.
Hence 0 and 1 are the roots of p(x) but p′(x) has no root.
If p = 3 then p(x) = x4 + x + 0 = x4 + x and p′(x) = 4x3 + 1 = x3 + 1.
Hence 0 and 2 are the roots of p(x) and 2 is als a root of p′(x).
If p = 5 then p(x) = x4 + x + 1 and p′(x) = 4x3 + 1.
Hence 1 is the only root of p′(x) but 1 is not a root of p(x).
If p = 7 then p′(x) = 4x3 + 1 has no root. Ans. 2.
20.
The graph of f intersects a line y = a parallel to X-axis if and only if there exists x in R such that f(x) = a. Hence Ans. 1, 2, 3, 4.
21.
If the given a > 2 then a can be written as in 1.
{Theorem. A positive integer n is a sum of two squares iff in the prime factorization of n the power of any prime factor p ≡ 3 (mod 4) of n is even.
i.e. n is a sum of two squares iff n factors as n = ab2 , where a has no prime factor p ≡ 3 (mod 4).
If the given a > 2 then by the above theorem, a = bc2, where b has no prime factor p ≡ 3 (mod 4), i.e all the prime factors of b are of the form p ≡ 1(mod 4). Hence, by Fermat′s theorem on the sum of two squares, all the prime factors of b are the sum of two squares.
Since the product of two sums of two squares is a sum of two squares, b = pd2, where p is one of the prime factors of b. Hence a = p(cd)2.}
Let a = 5 = 1 + 4. Then a cannot be written as in 2, 3 and 4, since the least prime of the form p ≡ 3 (mod 4) is 7 and that of p ≡ 1 (mod 4) is 5, in these cases pd2 or pqd2 are greater than 5, for any integer d. Ans. 2, 3, 4.
22.
Comments
Post a Comment