Algebra Practice Questions

The number of Abelian groups of order \( n = p_1^{a_1} \cdot p_2^{a_2} \cdots \) is \( p(a_1) \cdot p(a_2) \cdots \), where \( p \) is the partition function. For \( 2^6 \), we need partitions of 6: \( (6), (5,1), (4,2), (4,1,1), (3,3), (3,2,1), (3,1,1,1), (2,2,2), (2,2,1,1), (2,1,1,1,1), (1,1,1,1,1,1) \). So \( p(6) = 11 \). For \( 3^3 \), partitions of 3: \( (3), (2,1), (1,1,1) \). So \( p(3) = 3 \). Total groups \( = 11 \times 3 = 33 \).

Answer: (C) 33

The elementary divisors (prime powers) are: \( 8 = 2^3 \) (from \( \mathbb{Z}_8 \)) and \( 27 = 3^3 \) (from \( \mathbb{Z}_{27} \)). Since \( \gcd(8, 27) = 1 \), there is only one copy of each prime power. To form the largest invariant factor \( d_k \), we take the largest power of each prime: \( d_k = 8 \times 27 = 216 \). In fact, the group is cyclic: \( \mathbb{Z}_8 \times \mathbb{Z}_{27} \cong \mathbb{Z}_{216} \), and the single invariant factor is 216.

Answer: (B) 216

We calculate the order of each component: Order of \( 1 \) in \( \mathbb{Z}_4 \): \( 4/\gcd(1,4) = 4 \). Order of \( 4 \) in \( \mathbb{Z}_6 \): \( 6/\gcd(4,6) = 6/2 = 3 \). Order of \( 5 \) in \( \mathbb{Z}_{12} \): \( 12/\gcd(5,12) = 12/1 = 12 \). The order of the element in the direct product is \( \text{lcm}(4, 3, 12) = 12 \).

Answer: (C) 12

For cyclic groups, a homomorphism \( \phi: \mathbb{Z}_m \to \mathbb{Z}_n \) is completely determined by the image of a generator. The number of homomorphisms from \( \mathbb{Z}_m \) to \( \mathbb{Z}_n \) is \( \gcd(m, n) \). So the number of homomorphisms from \( \mathbb{Z}_{20} \) to \( \mathbb{Z}_{20} \) is \( \gcd(20, 20) = 20 \). Equivalently, the generator 1 can map to any element of \( \mathbb{Z}_{20} \) whose order divides 20 — which is every element, giving 20 homomorphisms.

Answer: (D) 20

Check the divisibility condition \( d_1 \mid d_2 \mid \cdots \mid d_k \) for each: (A) \( 3 \mid 12 \) and \( 12 \mid 60 \). OK. (B) \( 2 \mid 4 \), \( 4 \mid 8 \), \( 8 \mid 24 \). OK. (C) \( 5 \mid 10 \), \( 10 \mid 30 \). OK. (D) \( 4 \mid 6 \)? Since \( 6/4 = 1.5 \), this fails. So \( 4, 6, 12 \) cannot be invariant factors.

Answer: (D) \( 4, 6, 12 \)

By definition, \( \ker(\phi) = \{ A \in GL_2(\mathbb{R}) \mid \phi(A) = 1 \} = \{ A \in GL_2(\mathbb{R}) \mid \det(A) = 1 \} \). This is exactly the special linear group \( SL_2(\mathbb{R}) \). Note that (A) is incorrect because trace zero does not imply determinant 1, and (C) is the orthogonal group which has \( \det(A) = \pm 1 \), not necessarily 1.

Answer: (B) \( SL_2(\mathbb{R}) \)

We need \( (a, b) \) such that \( 2(a,b) = (0,0) \). For \( \mathbb{Z}_2 \): \( 2a \equiv 0 \pmod{2} \), so \( a \in \{0, 1\} \) (2 choices). For \( \mathbb{Z}_4 \): \( 2b \equiv 0 \pmod{4} \), so \( b \in \{0, 2\} \) (2 choices). Total elements satisfying \( 2(a,b) = (0,0) \) is \( 2 \times 2 = 4 \). We must subtract the identity \( (0,0) \) which has order 1. Elements of order exactly 2 = \( 4 - 1 = 3 \).

Answer: (C) 3

\( U(24) = \{ k : 1 \leq k < 24, \gcd(k, 24) = 1 \} \). The elements are \( \{1, 5, 7, 11, 13, 17, 19, 23\} \), so \( |U(24)| = 8 \). Note that \( a^2 \equiv 1 \pmod{24} \) for all \( a \in U(24) \) (e.g., \( 5^2 = 25 \equiv 1 \), \( 7^2 = 49 \equiv 1 \), etc.). Since every non-identity element has order 2, the group is elementary Abelian of order 8. Thus \( U(24) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \).

Answer: (C) \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \)

By the First Isomorphism Theorem, \( \mathbb{Z}_{48}/\ker(\phi) \cong \text{Im}(\phi) \). So \( |\text{Im}(\phi)| = 48 / |\ker(\phi)| \). The image must divide both 48 (as a quotient of \( \mathbb{Z}_{48} \)) and 36 (as a subgroup of \( \mathbb{Z}_{36} \)), so \( |\text{Im}(\phi)| \) divides \( \gcd(48, 36) = 12 \). If \( |\ker(\phi)| = 1 \), then \( |\text{Im}(\phi)| = 48 \), but \( 48 > 36 \), which is impossible. Thus, the kernel cannot be trivial. (Note: checking all options — \( |\ker| = 4 \) gives \( |\text{Im}| = 12 \), and \( 12 \mid 36 \), so this works. \( |\ker| = 2 \) gives \( |\text{Im}| = 24 \), and \( 24 \nmid 36 \), so this also cannot occur. \( |\ker| = 6 \) gives \( |\text{Im}| = 8 \), and \( 8 \nmid 36 \), so this also cannot occur.)

Answer: (D) 1

The primes involved are 3 and 5. Powers of 3: \( 9, 3 \). Powers of 5: \( 25, 5 \). We arrange them to satisfy the divisibility chain \( d_1 \mid d_2 \). Take the smallest power of each prime for \( d_1 \): \( d_1 = 3 \times 5 = 15 \). Take the largest power of each prime for \( d_2 \): \( d_2 = 9 \times 25 = 225 \). Check: \( 15 \mid 225 \) since \( 225/15 = 15 \). This holds. The invariant factors are \( 15, 225 \).

Answer: (A) \( 15, 225 \)

The order of the image (a subgroup of \( \mathbb{Z}_6 \)) must divide 6. Possible orders: 1, 2, 3, 6. By the First Isomorphism Theorem, \( \text{Im}(\phi) \cong \mathbb{Z}_{15}/\ker(\phi) \), so its order must also divide 15. The common divisors of 6 and 15 are 1 and 3. Thus, the maximum possible size is 3.

Answer: (B) 3

We require \( \phi(ab) = \phi(a)\phi(b) \). LHS: \( (ab)^2 = abab \). RHS: \( a^2 b^2 = aabb \). Thus, we need \( abab = aabb \), i.e., \( ba = ab \) for all \( a, b \in G \). This means elements commute. So \( \phi \) is a homomorphism if and only if \( G \) is Abelian.

Answer: (C) \( G \) is Abelian

The formula is \( \phi(x) = 3x \pmod{12} \). We want \( 3x \equiv 0 \pmod{12} \). This means \( 12 \mid 3x \), or equivalently \( 4 \mid x \). In \( \mathbb{Z}_{12} \), the multiples of 4 are \( \{0, 4, 8\} \).

Answer: (B) \( \{0, 4, 8\} \)

We know that \( \text{Inn}(G) \cong G/Z(G) \). Since \( \mathbb{Z}_{12} \) is Abelian, the center \( Z(G) = G \). Therefore, \( \text{Inn}(G) \cong G/G = \{e\} \). Alternatively, for any \( g \in G \), the inner automorphism \( \phi_g(x) = gxg^{-1} = x \) (since \( G \) is Abelian), so every inner automorphism is the identity map.

Answer: (D) The trivial group \( \{e\} \)

Finite groups cannot be isomorphic to their proper subgroups (by the pigeonhole principle/counting). This eliminates (A), (C), and (D). For \( \mathbb{Z} \), the map \( \phi: \mathbb{Z} \to 2\mathbb{Z} \) defined by \( \phi(n) = 2n \) is an isomorphism. \( 2\mathbb{Z} \) is a proper subgroup of \( \mathbb{Z} \), yet \( \mathbb{Z} \cong 2\mathbb{Z} \).

Answer: (B) \( \mathbb{Z} \)

To find the inverse, we solve \( e^x = y \) for \( x \). Taking the natural log of both sides: \( x = \ln(y) \). So \( \phi^{-1}(y) = \ln(y) \). This maps \( (\mathbb{R}^+, \cdot) \) back to \( (\mathbb{R}, +) \) and is indeed a homomorphism since \( \ln(ab) = \ln(a) + \ln(b) \).

Answer: (A) \( \phi^{-1}(y) = \ln(y) \)

The number of homomorphisms from \( \mathbb{Z}_m \) to \( \mathbb{Z}_n \) is \( \gcd(m, n) \). Here \( \gcd(10, 20) = 10 \). There are 10 such homomorphisms.

Answer: (C) 10

(A), (B), and (D) are standard properties of homomorphisms. (C) is false. The image is always a subgroup of \( H \), but it is not necessarily normal unless \( H \) is Abelian or other specific conditions are met. For example, the inclusion \( \phi: \langle (12) \rangle \hookrightarrow S_3 \) has image \( \{e, (12)\} \), which is not normal in \( S_3 \).

Answer: (C) The image \( \phi(G) \) is a normal subgroup of \( H \)

We calculate \( \phi_r(s) = rsr^{-1} \). In \( D_4 \), we use the relation \( rs = sr^{-1} \). So \( rsr^{-1} = (sr^{-1})r^{-1} = sr^{-2} \). Since \( r^{-2} = r^2 \) in \( D_4 \) (as \( r^4 = e \)), we get \( \phi_r(s) = sr^2 \).

Answer: (C) \( sr^2 \)

An automorphism of \( \mathbb{Z}_{12} \) is determined by where it sends the generator 1. The image must be another generator (a unit mod 12). So \( \text{Aut}(\mathbb{Z}_{12}) \cong U(12) \). Here \( U(12) = \{1, 5, 7, 11\} \). Notice \( 5^2 = 25 \equiv 1 \), \( 7^2 = 49 \equiv 1 \), \( 11^2 = 121 \equiv 1 \pmod{12} \). Since all non-identity elements have order 2, this is the Klein 4-group. So \( \text{Aut}(\mathbb{Z}_{12}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \).

Answer: (B) \( \mathbb{Z}_2 \times \mathbb{Z}_2 \)

The order of a permutation written in disjoint cycles is the least common multiple of the cycle lengths. The cycle lengths are 3 and 2. Order \( = \text{lcm}(3, 2) = 6 \). (The fact that it is in \( S_5 \) implies there is a fixed point, which has "length 1" and doesn't affect the lcm.)

Answer: (B) 6

First, identify \( H = \langle 4 \rangle = \{0, 4, 8\} \). The order of \( H \) is 3. The order of the quotient group is \( |G/H| = 12/3 = 4 \). Since \( \mathbb{Z}_{12} \) is cyclic, any quotient of \( \mathbb{Z}_{12} \) must also be cyclic. The unique cyclic group of order 4 is \( \mathbb{Z}_4 \).

Answer: (B) \( \mathbb{Z}_4 \)

Let \( n_5 \) be the number of Sylow 5-subgroups. By Sylow's Third Theorem: (1) \( n_5 \equiv 1 \pmod{5} \), and (2) \( n_5 \) must divide \( 20/5 = 4 \). The divisors of 4 are 1, 2, 4. Checking the congruence: \( 1 \equiv 1 \pmod{5} \) (Yes), \( 2 \not\equiv 1 \pmod{5} \), \( 4 \not\equiv 1 \pmod{5} \). Thus, \( n_5 = 1 \). (This implies the Sylow 5-subgroup is normal.)

Answer: (A) 1

A cycle of length \( k \) can be written as \( k-1 \) transpositions. A permutation is even if the total number of transpositions is even. (A) Length 2 → 1 transposition (Odd). (B) Length 4 → 3 transpositions (Odd). (C) Length 3 → 2 transpositions (Even). (D) Lengths 2+3+2 → \( 1+2+1 = 4 \) transpositions (Even). Both (C) and (D) are even, but (C) is the simplest canonical example of an even permutation.

Answer: (C) \( (1\ 2\ 3) \)

For \( D_n \) with \( n \geq 3 \): if \( n \) is even, the center is \( Z(D_n) = \{e, r^{n/2}\} \). If \( n \) is odd, \( Z(D_n) = \{e\} \). Here \( n = 4 \) (even), so \( Z(D_4) = \{e, r^2\} \). This is because \( r^2 \) is rotation by \( 180° \), which commutes with all rotations and all reflections.

Answer: (B) \( \{e, r^2\} \)

A standard theorem states that every group of order \( p^2 \) is Abelian. It is isomorphic to either \( \mathbb{Z}_{p^2} \) (cyclic) or \( \mathbb{Z}_p \times \mathbb{Z}_p \). Since it could be either, (A) and (B) are not always true. However, since it is Abelian, (C) is always true. Also, the center of a \( p \)-group is always non-trivial (size at least \( p \)), so (D) is false.

Answer: (C) \( G \) is Abelian

The number of generators of \( \mathbb{Z}_n \) is given by Euler's totient function \( \phi(n) \). \( 24 = 2^3 \cdot 3 \). \( \phi(24) = 24 \cdot (1 - 1/2)(1 - 1/3) = 24 \cdot 1/2 \cdot 2/3 = 8 \). The generators are the numbers coprime to 24: \( \{1, 5, 7, 11, 13, 17, 19, 23\} \).

Answer: (C) 8

The commutator subgroup \( G' \) is the smallest normal subgroup such that \( G/G' \) is Abelian. We know \( S_3/A_3 \cong \mathbb{Z}_2 \), which is Abelian. So \( G' \subseteq A_3 \). Since \( S_3 \) is not Abelian, \( G' \neq \{e\} \). Since \( A_3 \cong \mathbb{Z}_3 \) is simple (it has no proper non-trivial subgroups), \( G' \) must be all of \( A_3 \).

Answer: (B) \( A_3 \cong \mathbb{Z}_3 \)

The subgroups of order 4 in \( Q_8 = \{1, -1, i, -i, j, -j, k, -k\} \) are: (1) \( \langle i \rangle = \{1, -1, i, -i\} \), (2) \( \langle j \rangle = \{1, -1, j, -j\} \), (3) \( \langle k \rangle = \{1, -1, k, -k\} \). All three are cyclic of order 4. There are exactly 3 such subgroups. (Note: \( Q_8 \) has a unique subgroup of order 2, which is \( \{1, -1\} \).)

Answer: (B) 3

The image is \( \text{Im}(\phi) = 3\mathbb{Z} = \{\ldots, -6, -3, 0, 3, 6, \ldots\} \). The codomain is \( \mathbb{Z} \). The index is the number of cosets of \( 3\mathbb{Z} \) in \( \mathbb{Z} \). The cosets are \( 3\mathbb{Z},\ 1+3\mathbb{Z},\ 2+3\mathbb{Z} \). There are 3 distinct cosets. Thus, the index is 3.

Answer: (B) 3

A ring where every element is idempotent is called a Boolean ring. A standard result (which can be proven by expanding \((a+b)^2\)) shows that such rings are always commutative. They are not necessarily fields (e.g., \(\mathbb{Z}_2 \times \mathbb{Z}_2\)), do not necessarily have characteristic 2, and need not be finite.

Answer: B) \(R\) must be commutative.

An element \(a\) in a commutative ring is a zero divisor if there exists a non-zero \(b\) such that \(ab = 0\). For \(9 \in \mathbb{Z}_{12}\), we have \(9 \times 4 = 36 \equiv 0 \mod 12\). Since 4 is non-zero in \(\mathbb{Z}_{12}\), 9 is a zero divisor. The other numbers (5,7,11) are all relatively prime to 12 and are therefore units, not zero divisors.

Answer: C) 9

The set of nilpotent elements, called the nilradical, is an ideal. It's easy to show closure under addition (using the binomial theorem) and multiplication by any ring element. The set of units is not closed under addition. The set of zero divisors is generally not closed under addition. The set of idempotents is not closed under addition.

Answer: C) The set of all nilpotent elements

By definition, \(1_S\) satisfies \(1_S \cdot s = s \cdot 1_S = s\) for all \(s \in S\). In particular, \(1_S \cdot 1_S = 1_S\) within \(S\), so this equation also holds in \(R\). Thus, \(1_S\) is an idempotent of \(R\). It need not equal \(1_R\) (e.g., in \(\mathbb{Z}_6\), the subring \(\{0,3\}\) has unity 3, not 1).

Answer: B) \(1_S\) must be an idempotent element of \(R\).

The characteristic of a direct product is the least common multiple of the characteristics of the component rings. \(\mathbb{Z}_4\) has characteristic 4, and \(\mathbb{Z}_6\) has characteristic 6. \(\text{lcm}(4,6) = 12\).

Answer: C) 12

The two right ideals are \(\{0\}\) and \(R\) itself. For any non-zero element \(a \in R\), the right ideal generated by \(a\), denoted \(aR\), must equal \(R\). Therefore, there exists some \(r \in R\) such that \(ar = 1_R\), proving \(a\) has a right inverse. This is enough to prove \(R\) is a division ring. Commutativity is not guaranteed (e.g., quaternions). Finiteness is not required.

Answer: B) II only

The quotient \(\mathbb{Z}[x]/I \cong \mathbb{Z}_2\). Since \(\mathbb{Z}_2\) is a field, \(I\) must be a maximal ideal. It is not principal (it requires two generators). It is also prime because \(\mathbb{Z}_2\) is an integral domain.

Answer: B) \(I\) is a maximal ideal.

In this ring, \(6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\). The element 3 divides the product \((1+\sqrt{-5})(1-\sqrt{-5})\) but does not divide either factor, so 3 is not prime. It is irreducible because if \(3 = \alpha \beta\), then taking norms gives \(9 = N(\alpha)N(\beta)\), and one can show the only possibilities lead to one factor being a unit. This ring is the classic example of an integral domain that is not a UFD.

Answer: B) Irreducible but not prime

By definition, \(P\) is prime if whenever \(ab \in P\), then \(a \in P\) or \(b \in P\). This is equivalent to the quotient ring \(R/P\) having no zero divisors, i.e., being an integral domain. The definition via quotient rings is often easier to work with than the element-wise definition.

Answer: B) \(R/P\) is an integral domain.

The ideals of \(\mathbb{Z}_n\) are exactly the sets \(d\mathbb{Z}_n\) where \(d\) divides \(n\). Each divisor \(d\) of \(n\) corresponds to a distinct ideal. So the number of ideals equals the number of positive divisors of \(n\). The lattice of ideals of \(\mathbb{Z}_n\) is isomorphic to the lattice of divisors of \(n\).

Answer: C) The number of positive divisors of \(n\)

The standard theorem: an ideal \(M\) is maximal iff the quotient ring \(R/M\) is a field. While maximal ideals are always prime, the condition that \(R/M\) is an integral domain corresponds to \(M\) being prime, not maximal. Maximal ↔ quotient is a field. Prime ↔ quotient is an integral domain.

Answer: A) \(R/M\) is a field.

An ideal \((p)\) in \(\mathbb{Z}[i]\) is maximal if and only if \(p\) is a Gaussian prime. The rational prime 3 remains prime in \(\mathbb{Z}[i]\) (since \(x^2+1\) is irreducible mod 3), so \((3)\) is maximal. For 2, we have \(2 = -i(1+i)^2\), so \((2)\) is not prime. For 5, \(5 = (2+i)(2-i)\), so \((5)\) is not prime. In \(\mathbb{Z}[i]\), a rational prime \(p\) generates a maximal ideal iff \(p \equiv 3 \mod 4\).

Answer: B) \((3)\)

Evaluation at a point is always a ring homomorphism. The kernel consists of all polynomials with \(f(1)=0\), which is exactly the principal ideal generated by \((x-1)\). The image is \(\mathbb{Z}\) (since evaluating an integer polynomial at 1 gives an integer).

Answer: A) A ring homomorphism with kernel \((x-1)\).

The map \(\phi: \mathbb{R}[x] \to \mathbb{C}\) defined by \(\phi(f(x)) = f(i)\) is a surjective ring homomorphism with kernel \((x^2+1)\). By the First Isomorphism Theorem, \(\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}\). Quotienting by an irreducible polynomial over a field gives a field extension.

Answer: B) \(\mathbb{C}\)

A ring homomorphism \(f: \mathbb{Z}_m \to \mathbb{Z}_n\) must map the image of 1 to an idempotent \(r\) such that \(mr \equiv 0 \mod n\). In \(\mathbb{Z}_{30}\), we need idempotents \(r\) with \(20r \equiv 0 \mod 30\). The valid homomorphisms correspond to 2 such elements.

Answer: B) 2

The preimage of a prime ideal under a ring homomorphism is always prime. (A) is false: the image could be a subring of a field that is not itself a field. (B) is false: the image could have zero divisors. (D) is false: the image of a maximal ideal need not be maximal.

Answer: C) If \(I\) is a prime ideal in \(S\), then \(\phi^{-1}(I)\) is a prime ideal in \(R\).

The condition \(f(0) \equiv 0 \mod 2\) means the constant term of \(f\) is even. Such polynomials are exactly those in the ideal generated by 2 and x. The ideal (2) consists of polynomials with all coefficients even. The ideal (x) consists of polynomials with zero constant term. The kernel requires only the constant term to be even.

Answer: A) \((2, x)\)

In \(\mathbb{F}_2\), check if the polynomial has roots: \(f(0)=1\), \(f(1)=1+1+1=1\). No roots, so no linear factor. A cubic with no linear factor must be irreducible over a field. For polynomials of degree 2 or 3 over a field, irreducibility is equivalent to having no roots in that field.

Answer: B) Irreducible.

Total number of monic polynomials of degree 2 over \(\mathbb{F}_5\) is \(5^2 = 25\). The number of reducible monic quadratics: products of distinct linear factors: \(\binom{5}{2} = 10\), plus squares of linear factors: 5, total 15. So irreducible count = 25 - 15 = 10.

Answer: C) 10

If a cubic has a root \(a \in F\), then by the Factor Theorem, \((x-a)\) divides \(f(x)\), so \(f(x) = (x-a)g(x)\) where \(g(x)\) is quadratic. Thus \(f(x)\) is reducible. It need not split completely into linear factors (the quadratic factor might be irreducible over \(F\)).

Answer: C) \(f(x)\) is reducible.

Consider the quotient \(\mathbb{Z}[x]/(x^2+1, 5)\). First mod out by 5: \(\mathbb{Z}[x]/(5) \cong \mathbb{F}_5[x]\). In \(\mathbb{F}_5\), \(x^2+1\) has roots: \(2^2+1=5\equiv 0\). So \(x^2+1\) factors as \((x-2)(x-3)\) in \(\mathbb{F}_5[x]\). Thus the quotient is isomorphic to \(\mathbb{F}_5 \times \mathbb{F}_5\), which has zero divisors, so the ideal is not prime.

Answer: C) Not prime.

\(x^4 + 4\) can be factored using Sophie Germain's identity: \(x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)\), so it is reducible. \(x^4 - 4x^2 + 4 = (x^2-2)^2\), reducible. \(x^4 + 5x^2 + 6 = (x^2+2)(x^2+3)\), reducible. \(x^4 + 1\) is irreducible over \(\mathbb{Q}\) (it factors over \(\mathbb{R}\) but not \(\mathbb{Q}\)).

Answer: B) \(x^4 + 1\)

A PID with a unique maximal ideal is precisely a discrete valuation ring. It need not be a field (e.g., \(\mathbb{Z}_{(p)}\), the localization of \(\mathbb{Z}\) at a prime, is a PID with unique maximal ideal \((p)\) but is not a field). PID + local = DVR is a standard classification.

Answer: B) A discrete valuation ring (DVR).

\(\mathbb{Z}[\sqrt{-5}]\) is the classic example of an integral domain that is not a UFD (e.g., 6 = 2·3 = (1+√-5)(1-√-5) shows non-unique factorization). \(\mathbb{Z}[x]\) is a UFD. \(\mathbb{Z}[\sqrt{-2}]\) is a UFD (class number 1). \(\mathbb{F}_p[x,y]\) is a polynomial ring over a field, hence a UFD.

Answer: C) \(\mathbb{Z}[\sqrt{-5}]\)

Nilpotent elements are contained in every prime ideal, and since maximal ideals are prime, they are contained in every maximal ideal. Thus the nilradical (intersection of all prime ideals) is contained in the Jacobson radical. Units are never in any proper ideal. Idempotents and zero divisors need not be in the Jacobson radical.

Answer: A) Every nilpotent element.

This is the definition of a maximal ideal. If \(R/I\) is a field, then \(I\) is maximal. It is certainly prime (since fields are integral domains), but it's more than just prime. Field quotient ↔ maximal ideal. Integral domain quotient ↔ prime ideal.

Answer: B) Maximal.

A finite integral domain is always a field (a standard theorem). Since \(R\) has no zero divisors and is commutative with unity, it is an integral domain. Being finite, it is a field. It follows from the pigeonhole principle: multiplication by a non-zero element is injective, hence surjective.

Answer: A) A field.

If \(u\) and \(v\) are units with inverses \(u^{-1}\) and \(v^{-1}\), then \(uv\) has inverse \(v^{-1}u^{-1}\), so the product is a unit. (A) is false (e.g., in \(\mathbb{Z}\), 1 and -1 are units, but 1+(-1)=0 is not). (C) is true only in certain rings. (D) is false. The set of units forms a multiplicative group.

Answer: B) The product of two units is a unit.

\(R = \mathbb{Z}[\sqrt{2}]\) is the ring of integers of \(\mathbb{Q}(\sqrt{2})\). It is a Euclidean domain (with norm \(N(a+b\sqrt{2}) = |a^2 - 2b^2|\)), hence a PID, hence a UFD, hence an integral domain. It is not a field because elements like 2 are not units.

Answer: B) \(R\) is a Euclidean domain.

In \(\mathbb{Z}\), the only non-zero prime ideals are of the form \((p)\) where \(p\) is prime, and these are maximal. So \(\mathbb{Z}\) has dimension 1. In \(\mathbb{Z}[x]\), the ideal \((x)\) is prime but not maximal (\(\mathbb{Z}[x]/(x) \cong \mathbb{Z}\), an integral domain but not a field). So \(\mathbb{Z}[x]\) has dimension 2. Similarly, \(\mathbb{Q}[x,y]\) has dimension 2.

Answer: B) \(\mathbb{Z}\)