Your Answer : (Not Answered)

Correct Answer : A

The value that must work is 10.

There are 7 days of the week. By the pigeonhole principle, any group of 8 or more people must contain two born on the same day of the week, while a group of 7 or fewer can have everyone on a different day.

So the property is guaranteed only when \(n \ge 8\). Of the options, only 10 qualifies. With 7, 6 or 5 people each person can have a different day, so the property can fail.

Your Answer : (Not Answered)

Correct Answer : A

Only (I) is true.

(I) Any two 3-cycles are conjugate in \(S_{n}\) by some \(\sigma\). If \(\sigma\) is odd and \(n \ge 5\), there is a transposition \(\tau\) disjoint from the 3-cycle \(c\), and \(\sigma\tau\) is even and conjugates c to the same target, since \(\tau\) commutes with c. So all 3-cycles are conjugate in \(A_{n}\).

(II) is false: \(S_{n}\)-classes can split in \(A_{n}\). In \(A_{5}\), the 24 five-cycles form one class in \(S_{5}\) but split into two classes of 12 in \(A_{5}\).

Your Answer : (Not Answered)

Correct Answer : D

The correct statement is any finitely generated ideal of a Boolean ring is principal.

In a Boolean ring every element satisfies \(x^{2} = x\) and \(2x = 0\). For two generators, put \(c = a + b + ab\). Then \(ca = a + ab + ab = a\) and similarly \(cb = b\), so \(a, b \in (c)\), while \(c \in (a, b)\). Hence \((a, b) = (c)\), and by induction every finitely generated ideal is principal.

The others are false. K[x] has infinitely many maximal ideals, such as \((x)\) and \((x - 1)\), so it is not local. The additive group Q is divisible and has no maximal subgroup. And a ring without unity need not have a maximal ideal: the zero-multiplication ring on Q has ideals equal to its additive subgroups, and Q has no maximal subgroup.

Your Answer : (Not Answered)

Correct Answer : C

Both (I) and (II) are true.

(I) In an abelian group, HK is a subgroup, and it is finite of order \(mn/|H \cap K|\), a number divisible by both \(m\) and \(n\), hence by \(d = \text{lcm}(m, n)\). A finite abelian group has a subgroup of every order dividing its order (the converse of Lagrange holds for abelian groups), so HK, and therefore G, has a subgroup of order \(d\).

(II) Every group of order 15 is cyclic, hence abelian. By Sylow's theorems the number of Sylow 5-subgroups divides 3 and is 1 mod 5, so it is 1; the number of Sylow 3-subgroups divides 5 and is 1 mod 3, so it is 1. Both are normal and intersect trivially, so \(G \cong \mathbf{Z}_{3} \times \mathbf{Z}_{5} \cong \mathbf{Z}_{15}\).

So the options that deny either statement are wrong.

Your Answer : (Not Answered)

Correct Answer : A

The splitting field of \(x^{3} - 2\) over Q has degree 6.

The roots are \(\sqrt[3]{2}\), \(\omega\sqrt[3]{2}\) and \(\omega^{2}\sqrt[3]{2}\), where \(\omega\) is a primitive cube root of unity, so the splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).

\[ \begin{aligned} [\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] &= 3 \\ [\mathbf{Q}(\sqrt[3]{2}, \omega) : \mathbf{Q}(\sqrt[3]{2})] &= 2 \\ \text{total} &= 3 \times 2 = 6 \end{aligned} \]

The first degree is 3 because \(x^{3} - 2\) is irreducible (Eisenstein at 2). The second is 2 because \(\omega\) satisfies \(x^{2} + x + 1\) and is not real, while \(\mathbf{Q}(\sqrt[3]{2})\) is real. The Galois group is \(S_{3}\), of order 6. The answer 3 stops at the real root, and 2 and 4 do not divide in the right way.

Your Answer : (Not Answered)

Correct Answer : D

The coefficient is 0.

Every term of \((x^{2} + x^{3} + x^{4} + x^{5})^{5}\) is a product of five factors, each of degree at least 2, so every term has degree at least \(5 \times 2 = 10\). In fact

\[ \begin{aligned} (x^{2} + \cdots + x^{5})^{5} \\ = x^{10}(1 + x + x^{2} + x^{3})^{5} \end{aligned} \]

so the lowest power present is \(x^{10}\), and \(x^{8}\) does not occur.

13, 10 and 20 would be coefficients of higher powers; for example the coefficient of \(x^{11}\) is 5. The trap is to count solutions of \(e_{1} + \cdots + e_{5} = 8\) without noticing each \(e_{i} \ge 2\).

Your Answer : (Not Answered)

Correct Answer : B

The quadratic residue is 40.

\(40 \equiv -3 \pmod{43}\). Using Legendre symbols, with \(43 \equiv 3 \pmod 4\):

\[ \begin{aligned} \left(\tfrac{-1}{43}\right) &= -1 \\ \left(\tfrac{3}{43}\right) &= -\left(\tfrac{43}{3}\right) = -\left(\tfrac{1}{3}\right) = -1 \\ \left(\tfrac{-3}{43}\right) &= (-1)(-1) = 1 \end{aligned} \]

So 40 is a residue. (Indeed \(13^{2} = 169 = 3 \cdot 43 + 40\).) The others are not: -1 and 3 have symbol -1 as computed, and since \(43 \equiv 3 \pmod 8\), \((2/43) = -1\).

Your Answer : (Not Answered)

Correct Answer : B

The number that is not a sum of two squares is 1111.

A number is a sum of two squares exactly when every prime \(\equiv 3 \pmod 4\) in its factorisation appears to an even power. \(1111 = 11 \times 101\), and 11 \(\equiv 3 \pmod 4\) appears once. A quicker check: \(1111 \equiv 3 \pmod 4\), and a sum of two squares is 0, 1 or 2 mod 4.

The others are sums of two squares: \(405 = 18^{2} + 9^{2}\), \(117 = 9^{2} + 6^{2}\), and \(164 = 10^{2} + 8^{2}\).

Your Answer : (Not Answered)

Correct Answer : D

The order is \(lcm(l_{1}, \ldots, l_{k})\).

Disjoint cycles commute, so \(\sigma^{m} = \sigma_{1}^{m}\sigma_{2}^{m}\cdots\sigma_{k}^{m}\), and since they move different points, \(\sigma^{m}\) is the identity only if each \(\sigma_{i}^{m}\) is. A cycle of length \(l_{i}\) has order \(l_{i}\), so \(m\) must be a multiple of every \(l_{i}\); the least such \(m\) is the lcm.

For example, \((1\,2)(3\,4\,5)\) has order 6, which is not the sum 5 or the gcd 1; and \((1\,2)(3\,4)\) has order 2, not the product 4.

Your Answer : (Not Answered)

Correct Answer : D

The principal ideal is \((2x, 3x, x^{2})\).

Since \(3x - 2x = x\), the ideal contains \(x\); and each generator \(2x, 3x, x^{2}\) is a multiple of \(x\). So \((2x, 3x, x^{2}) = (x)\), which is principal.

The others are not principal. \((2, x)\) consists of polynomials with even constant term; a single generator would have to divide both 2 and \(x\), so it would be \(\pm 1\), but 1 is not in the ideal. The same argument works for \((3, x)\). And \((x, x^{2} - x, x + 4) = (x, 4)\), which fails for the same reason.

Your Answer : (Not Answered)

Correct Answer : D

The quadratic form is \(X^{2} - Y^{2}\).

For \(w \in \mathbf{C}\), multiplication by \(w = a + ib\) is the \(\mathbf{R}\)-linear map with matrix \(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\) on the basis 1, \(i\), so its trace is \(2a = 2\,\mathrm{Re}\,w\). Hence the trace form is \(B(x, y) = 2\,\mathrm{Re}(xy)\).

Its quadratic form at \(z = X + iY\) uses \(z^{2} = X^{2} - Y^{2} + 2iXY\):

\[ B(z, z) = 2\,\mathrm{Re}(z^{2}) = 2(X^{2} - Y^{2}) \]

With the usual convention \(q(z) = \tfrac{1}{2}B(z, z)\), this is \(X^{2} - Y^{2}\); with \(q = B(z, z)\) it differs only by the factor 2. Either way it is indefinite, of signature (1, 1), which rules out \(X^{2}\), \(Y^{2}\) and the positive definite \(X^{2} + Y^{2}\).

Your Answer : (Not Answered)

Correct Answer : B

The unit group is simple if and only if \(p = 3\).

\(\mathbf{F}_{p}^{\times}\) is cyclic of order \(p - 1\), and a cyclic group is simple exactly when its order is prime. So we need \(p - 1\) prime. Since \(p - 1\) is even for every odd prime, it can be prime only if \(p - 1 = 2\), that is \(p = 3\).

For \(p = 2\) the unit group is trivial, and the trivial group is not counted as simple. For \(p \ge 5\), \(p - 1\) is an even number at least 4, so the group has a proper non-trivial subgroup. Fermat primes \(2^{n} + 1\) with \(n \ge 2\) give \(p - 1 = 2^{n}\), again composite.

Your Answer : (Not Answered)

Correct Answer : D

The ring is not a UFD.

For an odd prime \(p\), \(1 + p\) is even, and

\[ 1 + p = (1 + \sqrt{-p})(1 - \sqrt{-p}) \]

The norm \(N(a + b\sqrt{-p}) = a^{2} + pb^{2}\) never equals 2 when \(p \ge 3\), so 2 is irreducible (its norm is 4 and any proper factor would have norm 2). But 2 divides the product without dividing either factor, since \((1 \pm \sqrt{-p})/2\) is not in the ring. An irreducible that is not prime means unique factorisation fails (for \(p = 3\): \(4 = 2 \cdot 2 = (1 + \sqrt{-3})(1 - \sqrt{-3})\)).

Every PID and every Euclidean domain is a UFD, so the other options are ruled out too.

Your Answer : (Not Answered)

Correct Answer : D

Q/H must be infinite.

(Q, +) is divisible: for every \(q\) and every positive integer \(n\) there is \(r\) with \(nr = q\). A quotient of a divisible group is divisible, so Q/H is divisible. It is also non-trivial because H is proper.

A non-trivial finite group of order \(m\) cannot be divisible: if \(g\) is not the identity, there is no \(x\) with \(mx = g\), since \(mx = 0\) for every element. So Q/H cannot be finite, of either parity.

For example, with H = Z the quotient Q/Z is infinite: the classes of \(1/n\) for \(n = 1, 2, 3, \ldots\) are all different.

Your Answer : (Not Answered)

Correct Answer : A

The sum is 24.

Gauss's identity says \(\sum_{d|n}\varphi(d) = n\) for every \(n\). Checking with the divisors of 24:

\[ \begin{aligned} &\varphi(1) + \varphi(2) + \varphi(3) + \varphi(4) \\ &+ \varphi(6) + \varphi(8) + \varphi(12) + \varphi(24) \\ &= 1 + 1 + 2 + 2 + 2 + 4 + 4 + 8 = 24 \end{aligned} \]

The reason is that each of the numbers 1 to 24 has a unique gcd \(g\) with 24, and those with gcd \(g\) number \(\varphi(24/g)\). 25 and 23 are off by one, and 20 omits terms.

Your Answer : (Not Answered)

Correct Answer : C

The possible value is \(n = 10\).

The dihedral group \(D_{5}\), the symmetries of a regular pentagon, has 10 elements and is non-abelian: a rotation and a reflection do not commute.

Each of the other orders forces every group to be abelian. A group of prime order 13 is cyclic. A group of order \(9 = 3^{2}\) has order \(p^{2}\), and every group of order \(p^{2}\) is abelian. A group of order \(15 = 3 \times 5\) is cyclic, because 3 does not divide \(5 - 1 = 4\), so the Sylow theorems leave only one Sylow subgroup of each order, both normal.

Your Answer : (Not Answered)

Correct Answer : D

The statement that need not be true is If F is infinite, then \(\mathbf{Z}_{p}\) cannot be embedded in F for any prime \(p\).

The field of rational functions \(\mathbf{Z}_{p}(x)\), or the algebraic closure of \(\mathbf{Z}_{p}\), is infinite but has characteristic \(p\) and contains \(\mathbf{Z}_{p}\).

The other three always hold. Every field contains its prime subring, which is \(\mathbf{Z}\) (characteristic 0) or \(\mathbf{Z}_{p}\) (characteristic \(p\)), giving A. A field has no zero divisors, and \(\mathbf{Z}_{n}\) for composite \(n\) has them, so it cannot sit inside F (B). And if \(\mathbf{Z}\) embeds, every nonzero integer is invertible in F, so all fractions \(a/b\) are there and Q embeds (C).

Your Answer : (Not Answered)

Correct Answer : A

The ring that is not Noetherian is \(\mathbf{R}[x_{1}, x_{2}, \ldots]\).

In the polynomial ring in infinitely many variables, the chain of ideals

\[ \begin{aligned} (x_{1}) &\subset (x_{1}, x_{2}) \\ &\subset (x_{1}, x_{2}, x_{3}) \subset \cdots \end{aligned} \]

is strictly increasing and never stops, so the ascending chain condition fails. The ideal generated by all the variables is not finitely generated.

\(\mathbf{Z}\) and \(\mathbf{Z}[i]\) are PIDs, hence Noetherian, and \(\mathbf{Z}_{205}[x]\) is Noetherian by Hilbert's basis theorem, since \(\mathbf{Z}_{205}\) is a finite ring and so Noetherian.

Your Answer : (Not Answered)

Correct Answer : A

Only one statement is true: \(\sum_{k=0}^{6} (-1)^{k}\binom{13}{2k} = -64\). The final key accepts this option alone.

Expand \((1 + i)^{13} = \sum_{j} \binom{13}{j} i^{j}\). The even terms \(j = 2k\) give \(i^{2k} = (-1)^{k}\), the real part; the odd terms give \(i^{2k+1} = (-1)^{k}i\), the imaginary part. In polar form:

\[ \begin{aligned} 1 + i &= \sqrt{2}\,e^{i\pi/4} \\ (1 + i)^{13} &= 2^{13/2}\,e^{13i\pi/4} \\ &= 64\sqrt{2}\left(-\tfrac{1}{\sqrt{2}} - \tfrac{i}{\sqrt{2}}\right) \\ &= -64 - 64i \end{aligned} \]

since \(\frac{13\pi}{4} = 2\pi + \frac{5\pi}{4}\). So the even alternating sum is \(-64\), not 64, and the odd alternating sum is also \(-64\), which is neither 128 nor \(-128\).

Your Answer : (Not Answered)

Correct Answer : B

The true statement is \(p(X)\) is reducible in the ring \(\mathbb{Q}[X]\).

Sophie Germain's identity factors it over the integers:

\[ \begin{aligned} X^{4} + 4 &= (X^{2} + 2)^{2} - 4X^{2} \\ &= (X^{2} - 2X + 2)(X^{2} + 2X + 2) \end{aligned} \]

The zeros are \(1 \pm i\) and \(-1 \pm i\). They all lie in \(\mathbb{Q}(i)\), so the splitting field is \(\mathbb{Q}(i)\), of degree 2 over \(\mathbb{Q}\), not 3 or 4. The zeros do not form a group under multiplication: 1 is not a zero, and \((1+i)(1-i) = 2\) is not a zero either.