Original public practice · AR-1 · October 2026

Abstract algebra II: rings and fields

Connect quotient structures, presentations, field extensions and ideal arithmetic in complete arguments. This form contains 10 multipart questions and 100 practice points.

This is undergraduate abstract algebra. It is separate from school Algebra I and II. These original problems sample the published material; they are not an official university paper, a calibrated grade prediction or an examination of every course objective.

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1. Rings and presentations

40 practice points.

Question 1: Reconstruct a finite algebra from two local observations

10 points

Work over F₅. Let R=F₅[x]/((x−1)²(x+2)). Write x for its residue class. Give ring arguments, not only dimension counts. Suggested time: 24 minutes.

(a) · 3 points

Construct an isomorphism R→F₅[ε]/(ε²)×F₅ using values and first-order coefficients. Prove its kernel and surjectivity explicitly.

(b) · 4 points

Determine all ideals of R, its maximal ideals, the number of units and the complete set of nilpotent elements. Prove your lists are exhaustive.

(c) · 3 points

Show e from part a is a nontrivial idempotent. Find the annihilator of e in R and explain why R cannot be a quotient of a field by a proper ideal.

Review question 1: worked argument and scoring criteria

(a)

Send p to (p(1)+p′(1)ε,p(−2)); formal division by (x−1)² gives the first component, so this is a ring map in characteristic 5. Its kernel is divisibility by both (x−1)² and x+2. They are coprime, hence the kernel is their product ideal. Put e=4(x−1)²: at −2 its value is 4·9=1 mod 5, and at 1 both its value and derivative vanish. Thus e maps to (0,1) and 1−e to (1,0). For any (a+bε,c), the polynomial (a+b(x−1))(1−e)+ce has that image. This proves surjectivity and the isomorphism.

1 point: Defines and justifies the two ring observations.
1 point: Proves the exact product kernel using coprimality.
1 point: Checks e and constructs a preimage for every target.

(b)

In D=F₅[ε]/ε², a+bε is a unit iff a≠0, with inverse a⁻¹−ba⁻²ε. Any ideal containing such an element is D; any nonzero proper ideal consists of multiples of ε and equals (ε), since a nonzero scalar coefficient is invertible. Thus D has exactly 0,(ε),D. A field has ideals 0,F₅. Every ideal in D×F₅ is a product ideal: multiply its elements by (1,0),(0,1) to isolate components. Hence R has six ideals. Maximal ideals correspond to (ε)×F₅ and D×0; their pullbacks are (x−1) and (x+2). Units number (4·5)·4=80. Nilpotents correspond exactly to (bε,0), b∈F₅: a nilpotent has zero field component and zero constant term in D, and conversely these square to zero. There are five, including zero.

1 point: Proves the three-ideal classification in the dual-number factor.
1 point: Proves product-ideal exhaustion and identifies the two maximal ideals.
1 point: Derives the unit count with an inverse criterion.
1 point: Characterizes all nilpotents in both directions.

(c)

Its image (0,1) squares to itself and is neither zero nor identity, so e is a nontrivial idempotent. Multiplication by (0,1) kills exactly D×0, the ideal generated by (1,0). Thus Ann(e)=(1−e)R. A field has only zero as a proper ideal, and its quotient by that ideal is a field. R is not a field because e(1−e)=0 with both factors nonzero. This supplies an obstruction rather than inferring it from the cardinality 125.

1 point: Checks nontrivial idempotence through the proved isomorphism.
1 point: Computes the exact annihilator with both containments.
1 point: Uses the proper-ideal structure of a field and a nonzero zero-divisor witness.

Targeted lesson returns:

Question 2: A polynomial quotient over the integers

10 points

Let I=(6,2t,t²)⊆ℤ[t], and let S=ℤ[t]/I. For n≥0 define Lₙ to be the set of coefficients of tⁿ in elements of I having degree at most n; include the zero polynomial. Suggested time: 24 minutes.

(a) · 4 points

Prove every class has a unique form a+bt with a modulo 6 and b modulo 2. Determine its multiplication, all units and all prime ideals.

(b) · 3 points

Compute every Lₙ and explain how lifts of their generators generate I without any division by a nonunit.

(c) · 3 points

Prove I is not principal, even though ℤ[t] is Noetherian and a UFD. Identify the exact missing implication in “finite ideal generation implies one generator.”

Review question 2: worked argument and scoring criteria

(a)

Every higher power vanishes, while 6 and 2t vanish. Conversely a polynomial belongs to I exactly when its constant coefficient is divisible by 6 and its linear coefficient by 2; all terms of degree ≥2 are unrestricted. This proves uniqueness and |S|=12. Multiplication is (a,b)(c,d)=(ac mod 6,ad+bc mod 2). A unit must have a unit constant term modulo 6. Conversely a=1 or 5 is odd and invertible, and the nilpotent term can be inverted by the finite expression a⁻¹−ba⁻²t. There are four units. Every prime ideal contains t since t²=0. Primes therefore correspond to prime ideals of S/(t)=ℤ/6ℤ, namely (2),(3), with field quotients F₂,F₃. The primes of S are (2,t),(3,t), both maximal.

1 point: Proves unique normal forms by both ideal-membership directions.
1 point: Derives the product law.
1 point: Proves the unit criterion and count.
1 point: Exhausts primes using nilpotence and quotient correspondence.

(b)

L₀=6ℤ, L₁=2ℤ and Lₙ=ℤ for n≥2, directly from the coefficient membership condition. The lifts 6,2t,t² have those leading coefficients. Given f∈I, subtract its constant coefficient divided by 6 times 6 and its linear coefficient divided by 2 times 2t; the divisions are divisions of known integer multiples, producing integers. The remaining polynomial is t² times an integer polynomial. Alternatively cancel highest coefficients with t² and its shifts, then finish the two low degrees. This never divides an arbitrary ring element by 2 or 6.

1 point: Computes all three ranges of coefficient ideals.
1 point: Gives a complete terminating reduction by the lifted generators.
1 point: Distinguishes known integer divisibility from a unit inverse.

(c)

If I=(g), then g divides the nonzero constant 6 in the domain ℤ[t], so degree additivity forces g to be a nonzero integer. Since g also divides t², its integer value divides the coefficient 1, hence g=±1. That would give I=ℤ[t], contradicted by evaluation t↦0 modulo 2, which kills I but not 1. Thus I is nonprincipal. Noetherian means every ideal has some finite generating list; neither that definition nor unique factorization of elements reduces all such lists to one generator. The present ideal is a concrete counterexample.

1 point: Uses degree and divisibility of 6 to restrict a hypothetical generator.
1 point: Uses t² and a properness witness to obtain a contradiction.
1 point: Separates Noetherian/UFD conditions from the PID property.

Targeted lesson returns:

Question 3: One presentation, every integer right-hand side

10 points

Let A:ℤ³→ℤ² have rows (4,6,2) and (2,8,4). Relations are columns. All operations must preserve integer lattices. Suggested time: 24 minutes.

(a) · 3 points

Construct an explicit unimodular column matrix V with AV=[[2,0,0],[0,2,0]]. Deduce the cokernel and give an independent determinantal-divisor check.

(b) · 3 points

For arbitrary b=(b₁,b₂)∈ℤ², classify consistency and give every integer solution of Ax=b with no omitted congruence conditions.

(c) · 4 points

Prove for an arbitrary integer matrix M that unimodular row and column changes preserve its presented abelian group, explaining why their roles differ. Diagnose replacing A by A/2 when computing its cokernel.

Review question 3: worked argument and scoring criteria

(a)

Take V with columns (2,−2,3), (−3,4,−6), (−2,3,−5). Multiplication gives the claimed AV, and det V=−1, so this is an allowed integer basis change. The cokernel is C₂⊕C₂. Independently the gcd of the entries of A is 2, and the three 2-by-2 minors are 20,12,8, with gcd 4. Thus Smith invariants are d₁=2,d₂=4/2=2, agreeing with the certificate.

1 point: Supplies and verifies the exact matrix product.
1 point: Checks unimodularity and reads the cokernel.
1 point: Checks entries/minors and recovers the same invariant factors.

(b)

The image is exactly 2ℤ² by the certificate, so consistency means both b entries even. Put cᵢ=bᵢ/2 and d=2c₂−c₁. Every solution is x=(c₂−2d−2r,2d+3r,−3d−5r), r∈ℤ. Indeed after dividing the equations by 2, eliminate x₁ to get 5x₂+3x₃=d. One particular solution is (2d,−3d) in these last coordinates; all differences satisfy 5y+3z=0, giving y=3r,z=−5r by coprimality. Substitution gives x₁ and proves exhaustiveness.

1 point: States and proves the exact parity consistency condition.
1 point: Gives a correct parameterized family.
1 point: Proves every solution lies in the family using the integer homogeneous equation.

(c)

For U∈GL_m(ℤ), V∈GL_n(ℤ), the column image of MV equals that of M because Vℤⁿ=ℤⁿ. The map y+im M↦Uy+im(UMV) is a well-defined isomorphism of cokernels, with inverse induced by U⁻¹. Thus V changes the relation basis in the source and U changes the ambient generator basis in the target; neither changes the group type. An integer matrix is unimodular precisely when its determinant is ±1: necessity follows by determinants of an integer inverse; sufficiency follows by the adjugate. Dividing A by 2 is not such an operation. It gives a surjection ℤ³→ℤ² here, with zero cokernel, whereas A has cokernel C₂². Dividing equations to solve an already even right-hand side is a different, justified calculation.

1 point: Explains source-image invariance under V.
1 point: Constructs both quotient maps for U.
1 point: Proves the determinant/integer-inverse criterion.
1 point: Computes and explains the failed division change, distinguishing equation solving.

Targeted lesson returns:

Question 4: Equal characteristic polynomials, different polynomial modules

10 points

Over ℚ put p(t)=t²+2, C=[[0,−2],[1,0]], S=diag(C,C), and T=[[C,I₂],[0,C]]. Regard each ℚ⁴ as a ℚ[t]-module by letting t act as its operator. Suggested time: 24 minutes.

(a) · 3 points

Determine the characteristic polynomial, minimal polynomial and invariant factors of S and T. Prove they are not similar without relying on eigenvalue pictures over ℂ.

(b) · 3 points

Prove e₄ is cyclic for T and construct the matrix of T in (e₄,Te₄,T²e₄,T³e₄). Explain what happens to ker p(T), compared with ker p(S).

(c) · 4 points

For any operator U on ℚⁿ prove ℚ[t]ⁿ/(tI−U)ℚ[t]ⁿ≅V_U by an explicit evaluation map and its full kernel. Explain why full rank of tI−U over ℚ(t) does not make this quotient zero.

Review question 4: worked argument and scoring criteria

(a)

Both block upper triangular matrices have characteristic polynomial p². We have p(S)=0, and p is irreducible over ℚ because a rational square cannot be −2; thus the minimal polynomial of S is p, and its module is two copies of ℚ[t]/p, with invariant factors p,p. For T, direct block multiplication gives p(T)=[[0,2C],[0,0]], nonzero of rank 2, and p(T)²=0. Its minimal polynomial divides p² but not p; irreducibility leaves p². Since this has degree 4, T is cyclic and has the single nonunit invariant factor p². Different annihilating minimal polynomials exclude similarity.

1 point: Computes the common characteristic polynomial and S module data.
1 point: Computes p(T) and proves the minimal polynomial is p².
1 point: Deduces invariant factors and a valid nonsimilarity obstruction.

(b)

We have p(T)e₄=(−4,0,0,0)≠0. The monic annihilator of e₄ divides p²; since p is irreducible and neither 1 nor p annihilates it, the annihilator is p². Hence no relation of degree below 4 among its iterates exists, so they form a basis. Since p²=t⁴+4t²+4, the matrix in this basis is the companion with columns e₂,e₃,e₄,(−4,0,−4,0), namely rows (0,0,0,−4),(1,0,0,0),(0,1,0,−4),(0,0,1,0). Since C is invertible, p(T) has kernel the first ℚ² block, dimension 2; p(S)=0 has kernel dimension 4.

1 point: Proves cyclicity from the vector annihilator, not just an assertion about T.
1 point: Constructs the full companion matrix with the correct signs.
1 point: Computes and contrasts the two polynomial kernels.

(c)

Define E(Σtᵏvₖ)=ΣUᵏvₖ. It is ℚ[t]-linear into V_U and onto via constants. E((tI−U)q)=0. For k≥1, tᵏv−Uᵏv=(tI−U)Σⱼ₌₀ᵏ⁻¹tᵏ⁻¹⁻ʲUʲv by telescoping. Summing gives q−E(q) in the image; if E(q)=0 this proves the reverse kernel inclusion. The induced map is therefore an isomorphism. The determinant det(tI−U) is nonzero monic, giving full rank over the fraction field, but it need not be a polynomial unit. The quotient over ℚ[t] is torsion and can be nonzero, with finite ℚ-dimension n; passing to ℚ(t) kills that torsion. For the present matrices its dimension is 4, not zero.

1 point: Defines the correct module evaluation and proves surjectivity.
1 point: Checks the image is killed and states a correct telescoping identity.
1 point: Uses it for the full kernel and quotient isomorphism.
1 point: Distinguishes polynomial units/fraction-field rank from a nonzero torsion quotient.

Targeted lesson returns:

2. Fields and Galois reasoning

30 practice points.

Question 5: One generator, four embeddings and three fixed fields

10 points

Let K=ℚ(√2,√7)⊆ℝ and α=√2+√7. Use rational algebra and explicit field maps; do not decide degrees from decimal approximations. Suggested time: 24 minutes.

(a) · 3 points

Prove [K:ℚ]=4 and K=ℚ(α). Determine and prove the minimal polynomial of α.

(b) · 3 points

List every ℚ-embedding of K into ℂ, prove the list exhaustive, and show K/ℚ is Galois with group C₂×C₂.

(c) · 4 points

Determine all proper intermediate fields other than ℚ, including the fixed field of simultaneous sign reversal. Prove these are exhaustive, and explain why α itself lies in none of them.

Review question 5: worked argument and scoring criteria

(a)

The number √7 is not a+b√2 with a,b∈ℚ: squaring would force ab=0, then either a²=7 or b²=7/2, both impossible by prime valuations of rational squares. Thus the second quadratic extension is genuine and [K:ℚ]=4. Since (√7−√2)α=5, we recover √2=(α−5/α)/2 and √7=(α+5/α)/2; α≠0, so K=ℚ(α). Squaring α gives α²=9+2√14 and hence (α²−9)²=56, or α⁴−18α²+25=0. Its degree must be 4 by the field equality, proving that this monic quartic is the minimal polynomial.

1 point: Proves √7 is absent from the first quadratic field.
1 point: Recovers both radicals from α with a justified nonzero denominator.
1 point: Derives the polynomial and proves minimality through the degree.

(b)

An embedding must send √2 to ±√2 and √7 to ±√7. Each independent sign choice defines an automorphism: use the basis 1,√2,√7,√14 and its multiplication rules to check the map, with the same sign change as inverse. Thus there are exactly four embeddings and all land in K. Equivalently K is the splitting field of (t²−2)(t²−7), which is separable in characteristic zero. The independent sign changes commute and square to identity; their four distinct maps form C₂×C₂. Exhaustiveness comes from the images of the two field generators, not from merely counting constructed maps.

1 point: Constructs all four embeddings with well-definedness or basis verification.
1 point: Proves there can be no others and checks normality/separability.
1 point: Establishes the group law and abstract identification.

(c)

Write x=a+b√2+c√7+d√14. Fixing x under the flip of √2 forces b=d=0, giving ℚ(√7); the flip of √7 gives ℚ(√2); simultaneous reversal forces b=c=0, giving ℚ(√14). The group has exactly three subgroups of order two, and by Galois correspondence these are all nontrivial proper intermediate fields. Each displayed field has degree two; no other intermediate degree is possible by the tower law, and the correspondence exhausts the fields of that degree. Since ℚ(α)=K has degree four, α belongs to none of them. Explicitly its nonzero √2 and √7 coefficients violate each of the corresponding fixed conditions.

1 point: Computes the simultaneous-reversal fixed field from coefficients.
1 point: Computes the other two fixed fields.
1 point: Uses the proved Galois correspondence for exhaustive classification.
1 point: Excludes α with a degree or explicit fixed-vector argument.

Targeted lesson returns:

Question 6: A cubic splitting field and a compatible radical expression

10 points

Let f(t)=t³−2t+2∈ℚ[t], let α be its real root, and let L be its splitting field in ℂ. Every radical expression in this question must include its choice or compatibility condition. Suggested time: 24 minutes.

(a) · 3 points

Prove f is irreducible, compute its discriminant, and identify Gal(L/ℚ). Determine the quadratic fixed field of its alternating subgroup.

(b) · 3 points

Prove L=ℚ(α,√−19), give [L:ℚ], and express the other roots using α and a chosen square root Δ of the discriminant. Prove f′(α) is nonzero.

(c) · 4 points

Construct a real expression for α using one square root and two real cube roots, with their product checked. Relate this to solvability of S₃ and explain why arbitrarily chosen complex cube-root branches need not give a root.

Review question 6: worked argument and scoring criteria

(a)

The rational root candidates ±1,±2 give values 1,3,6,−2, none zero, so the cubic is irreducible. For t³+pt+q the proved discriminant formula gives −4(−2)³−27·2²=−76. It is nonzero and not a rational square. Irreducibility makes the Galois group transitive on three roots, so it is A₃ or S₃; the nonsquare discriminant excludes A₃. Thus it is S₃. If Δ is the Vandermonde product, Δ²=−76 and the sign action has kernel A₃, giving its fixed field ℚ(Δ)=ℚ(√−19). This uses the characteristic-zero separable sign theorem.

1 point: Proves irreducibility using all rational candidates.
1 point: Computes the discriminant and correctly applies the transitive/sign criteria.
1 point: Identifies and justifies the quadratic fixed field.

(b)

Separability gives f′(α)≠0; alternatively gcd(f,f′)=1 follows from the nonzero discriminant. Let the other roots be β,γ and choose the sign of Δ so Δ=f′(α)(β−γ). This is possible since squaring the right side gives the discriminant. Since β+γ=−α, they equal (−α±Δ/f′(α))/2. Therefore all roots lie in ℚ(α,Δ), and conversely Δ as a product of root differences lies in L, proving equality. The real field ℚ(α) does not contain √−19; adjoining it has degree two. Hence [L:ℚ]=2·3=6, also matching |S₃|.

1 point: Proves the nonzero derivative and the squared-difference identity.
1 point: Derives the two remaining roots with a consistent square-root sign.
1 point: Proves both field containments and the degree.

(c)

Take u to be the real cube root of −1+√(19/27), and v the real cube root of −1−√(19/27). Their cubes sum to −2 and multiply to 1−19/27=8/27. Both u,v are real, so uv is the real cube root of 8/27, namely 2/3. Thus (u+v)³=u³+v³+3uv(u+v)=−2+2(u+v), proving α=u+v is a real root. The discriminant is negative, so there is only one real root; alternatively the other two roots from part b have nonreal imaginary difference and real sum, while a cubic cannot have two distinct real roots without all three being real. S₃ has the normal series 1⊲A₃⊲S₃ with cyclic factors of orders 3 and 2, so is solvable, consistent with the radical expression. Multiplying independently chosen complex branches by cube roots of unity can change uv; the cancellation requires uv=2/3. The expression has to enforce this compatibility rather than merely supply the two cubic equations.

1 point: Constructs the square-root and cube-root values with correct signs.
1 point: Verifies uv=2/3 and substitutes to prove the cubic equation.
1 point: Justifies the real root identification and the solvable-group statement.
1 point: Explains the precise branch compatibility and its role in cancellation.

Targeted lesson returns:

Question 7: Splitting is not the same as separability

10 points

Let F=F₃(u), where u is transcendental, and compare f(X)=X³−u with g(X)=X³−X−u. A pole at infinity of a rational function P(u)/Q(u) in reduced form has order deg P−deg Q when this is positive. Suggested time: 24 minutes.

(a) · 3 points

Prove f is irreducible over F. If θ³=u, determine the degree, roots, and number of F-embeddings of F(θ) into an algebraic closure. Is this a Galois extension?

(b) · 3 points

Prove g has no root in F using pole order at infinity. Deduce irreducibility and construct its full splitting field from one root η.

(c) · 4 points

Determine the Galois group of F(η)/F and its intermediate fields. Compare the fixed-field and automorphism conclusions with F(θ)/F, explaining why a Galois-correspondence argument cannot be transferred to the first extension.

Review question 7: worked argument and scoring criteria

(a)

A rational cube has u-adic valuation divisible by 3, since valuation is three times that of its cube root; u has valuation 1. Thus u is not a cube in F. A reducible cubic over a field has a root, so X³−u is irreducible and [F(θ):F]=3. In characteristic three, X³−u=(X−θ)³ in F(θ)[X]. It has only the one distinct root, and every F-embedding must send θ to that root, so only the identity embedding exists. The extension is the splitting field but is inseparable (derivative zero), hence not Galois. Splitting alone does not supply separability.

1 point: Proves the cube obstruction and cubic irreducibility.
1 point: Determines degree and the repeated-root factorization.
1 point: Counts embeddings and diagnoses the failed Galois condition.

(b)

If h∈F has a pole at infinity of order m>0, then h³−h has pole order 3m: the leading orders differ, so the highest pole cannot cancel. It cannot equal u, which has pole order 1. If h has no pole at infinity, neither h³ nor h nor their difference has such a pole, again excluding u. Thus g has no rational-function root and, being cubic, is irreducible. Its derivative is −1, so it is separable. For each c∈F₃, (η+c)³−(η+c)=η³−η because c³=c. The three distinct roots are η,η+1,η−1, all in F(η), so that degree-three field is the splitting field.

1 point: Proves the pole-order contradiction in both cases.
1 point: Deduces cubic irreducibility and separability.
1 point: Constructs all three roots and proves splitting in F(η).

(c)

The maps η↦η+c for c∈F₃ extend to F-automorphisms through the irreducible polynomial quotient, and all roots are in F(η). Composition adds c, so the group is cyclic of order 3. This extension is normal and separable; Galois correspondence gives no proper intermediate fields because C₃ has no nontrivial proper subgroups. The degree-three tower law also excludes intermediate degrees for either extension. In F(θ), however, the automorphism group is trivial, so its fixed field is the entire F(θ), not F. Applying the Galois fixed-field identity there would be false. Its absence of intermediate fields follows from degree, not from a correspondence with its one-element automorphism group. This distinguishes coincident field counts from valid theorem hypotheses.

1 point: Constructs all automorphisms and checks the group law.
1 point: Applies correspondence to the separable splitting extension correctly.
1 point: Computes the fixed field of the trivial automorphism group in the inseparable case.
1 point: Separates the tower-law conclusion from the invalid correspondence argument.

Targeted lesson returns:

3. Ideal arithmetic and synthesis

30 practice points.

Question 8: Two prime ideals and one ideal class in a changed field

10 points

Put s=√−6 and O=ℤ[s], the full integer ring because −6 is squarefree and is 2 modulo 4. Let I=(2,s) and J=(3,s). Results proved in earlier parts may be reused, but do not assume the class group has already been computed. Suggested time: 24 minutes.

(a) · 3 points

Give complete integer lattice bases and membership tests for I,J. Construct their quotient fields and prove both ideals maximal.

(b) · 4 points

Prove I²=(2), J²=(3), and IJ=(s) by both containments and explicit generator combinations. Deduce fractional inverses.

(c) · 3 points

Prove I and J are nonprincipal, and determine their classes relative to one another. State exactly what this proves and what remains before a full class-number claim.

Review question 8: worked argument and scoring criteria

(a)

The integer spans have coordinates (2m,n) and (3m,n), hence memberships a even and a divisible by 3 for a+bs. They are ideals: s·2=2s and s²=−6 lie in the first span, and s·3=3s and −6 lie in the second. Thus these spans equal the corresponding O-ideals. Evaluation s↦0 gives onto maps O→F₂ and O→F₃ because −6 vanishes in both fields. Their kernels are exactly these membership conditions. Therefore N(I)=2,N(J)=3, and both are maximal and prime.

1 point: Proves lattice membership and closure, not only the generator list.
1 point: Checks both quotient homomorphisms and kernels.
1 point: Derives norms and maximality from the fields.

(b)

I² has generators 4,2s,−6, all in 2O; 2=2·4−6 gives the reverse containment. J² has generators 9,3s,−6, all in 3O; 3=9−6 gives the reverse containment. IJ has generators 6,2s,3s,−6, all divisible by s since 6=−s², and s=3s−2s belongs to the product. Hence IJ=(s). These yield I⁻¹=I/2 and J⁻¹=J/3 by multiplying, and also I⁻¹=J/s and J⁻¹=I/s. The inverse identities concern fractional ideals, so denominators do not need to stay in O.

1 point: Verifies both containments for I².
1 point: Verifies both containments for J².
1 point: Verifies both containments for IJ.
1 point: Derives and checks the fractional inverse formulas.

(c)

A generator α=a+bs of an integral principal ideal must itself lie in O. Its index is |N(α)|=a²+6b². This cannot be 2 or 3: if b≠0 it is at least 6, and if b=0 neither is an integer square. Thus I,J are nonprincipal. Their squares are principal, so each class has order exactly two. Since IJ is principal, [J]=[I]⁻¹=[I]. This exhibits one nontrivial class and a subgroup C₂. It does not rule out other classes; an exhaustive representative bound and small-index classification are still needed.

1 point: Uses principal indices with the integral-generator justification.
1 point: Proves both norm obstructions and the class identities.
1 point: Distinguishes an exhibited subgroup from full class-group exhaustion.

Targeted lesson returns:

Question 9: From inverse certificates to global indices

10 points

Use nonzero integral ideals of a full quadratic integer ring, where the proved Dedekind theorem gives invertibility and finite prime factorization. For concrete calculations return to O=ℤ[√−6], I=(2,s), J=(3,s) from the preceding question; their stated products may be used even if you did not finish that question. Suggested time: 24 minutes.

(a) · 4 points

Prove A/PA is one-dimensional over O/P for invertible A and nonzero prime P. Use a finite inverse certificate, including the exact coefficient kernel; do not simply cancel A in a quotient.

(b) · 3 points

Using a prime factorization of B, prove N(AB)=N(A)N(B) including B=O. Apply the proof to compute [I:IJ], giving representatives and identifying a valid quotient generator.

(c) · 3 points

Describe O/(6) as a product of explicit rings, and determine its size and whether it has nonzero nilpotents. Prove the required CRT surjectivity rather than citing cardinalities alone.

Review question 9: worked argument and scoring criteria

(a)

Write 1=Σaᵢbᵢ with aᵢ∈A,bᵢ∈A⁻¹. Some ab is outside P. Choose c∈O with cab≡1 mod P, possible since O/P is a field. For x∈A, bx∈O and x−c(bx)a=(1−cab)x∈PA, proving generation by a+PA. The map O→A/PA, r↦ra+PA, has kernel P: inclusion is immediate in one direction; if ra∈PA then rab∈P since bA⊆O, and primality with ab∉P gives r∈P. Thus it induces an O/P-module isomorphism, proving dimension one. This is not an assumed ring isomorphism with an identity on A/PA.

1 point: Extracts a product outside P from a finite inverse certificate.
1 point: Proves quotient generation with an integral coefficient.
1 point: Proves both directions of the exact kernel statement.
1 point: Identifies the module isomorphism and its field hypotheses.

(b)

For B=P₁⋯Pᵣ, the chain A⊇AP₁⊇⋯⊇AB has each successive index N(Pⱼ) by part a. Finite coset representative multiplication gives [A:AB]=∏N(Pⱼ). The same chain starting at O gives N(B) equal to this product, and [O:AB]=[O:A][A:AB] proves the formula. B=O is the empty chain/product 1. For A=I,P=J, IJ=(s). The elements of I have even constant coordinate; (s) consists exactly of those with constant coordinate divisible by 6 and arbitrary s-coordinate, since s(a+bs)=−6b+as. Hence I/IJ has representatives 0,2,4, generated by 2 over F₃ and index 3. In the inverse certificate one may take a=2,b=1∈I⁻¹ and c=2, since 2ab=4≡1 mod J.

1 point: Proves the chain product and handles the empty case.
1 point: Derives multiplicativity from two correct chains.
1 point: Checks concrete representatives, generator and certificate coefficients.

(c)

The ideals 2O and 3O are comaximal because 3−2=1. The map O→O/2O×O/3O has kernel 6O: intersection equals product for comaximal ideals, by multiplying x in the intersection by a decomposition of 1. Given residues represented by a,b, the element 3a+4b has the chosen residues modulo 2 and 3 respectively. Since s²=−6, the two factors are F₂[z]/z² and F₃[z]/z², of sizes 4 and 9. Thus the product has 36 elements, agreeing with the principal norm of 6. The image of s is a nonzero pair of nilpotents, and s²=0 modulo 6; so this ring is not a product of fields. The direct-product description retains multiplication, not only the additive cardinality.

1 point: Proves the CRT kernel and explicit preimage formula.
1 point: Identifies both factor rings and the total size.
1 point: Checks a nonzero nilpotent and distinguishes the result from a product of fields.

Targeted lesson returns:

Question 10: Turn a geometric bound into an exhaustive answer

10 points

Continue with O=ℤ[√−6] embedded in ℂ by a+bs↦(a,b√6). You may use ideal invertibility, norm multiplicativity and the ideal identities from question 8 (Two prime ideals and one ideal class in a changed field). Do not assume a class-number table. Ordinary measurable planar area, translation invariance and finite additivity are available. Suggested time: 24 minutes.

(a) · 4 points

Prove that a planar lattice Λ of covolume d has a nonzero vector of squared length at most 4d/π. Include the half-open-cell area argument, convex symmetry step and passage from strict area to the endpoint.

(b) · 3 points

Deduce that every ideal class of O has an integral representative of index at most 3. Explain the covolume, the inverse-class direction and the numerical cutoff without decimals.

(c) · 3 points

Classify all ideals of index 2 or 3 in O and conclude the entire ideal class group. Explain why this conclusion could not have been drawn from the product identities alone.

Review question 10: worked argument and scoring criteria

(a)

Write Λ=Bℤ² and D=B[0,1)². Its translates partition the plane by integer/fractional parts. For a bounded measurable S, only finitely many translates meet S: the relevant lattice vectors lie in bounded S−D and have bounded integer inverse coordinates. The multiplicity m(x)=Σλ1_S(x+λ) on D therefore has integral area(S). If area(S)>d, some x has multiplicity at least two, giving distinct points of S whose nonzero difference is in Λ. Apply this with S=C/2 for a centered symmetric convex body C of area>4d; symmetry and convexity place the difference in C. For disks use radii r₀+1/n, where r₀²=4d/π. The obtained lattice points lie in one bounded disk, which contains finitely many lattice points by inverse coordinates. One nonzero point recurs infinitely often and has length ≤r₀. This proves the non-strict endpoint rather than applying the strict theorem at equality.

1 point: Proves exact half-open tiling and finite overlap.
1 point: Uses the area multiplicity identity to obtain a nonzero lattice difference.
1 point: Uses symmetry/convexity to turn the difference into a point in C.
1 point: Proves bounded lattice finiteness and the endpoint argument.

(b)

An integral ideal A with integer basis matrix B has covolume √6|det B|=√6N(A). Part a gives 0≠α∈A with |α|²≤(4√6/π)N(A). Then H=(α)A⁻¹ is integral, since α∈A, and AH=(α). Norm multiplicativity gives N(H)=|α|²/N(A)≤4√6/π<4: use √6<5/2 and π>3 to bound it by 10/3<4. Therefore the positive integer N(H) is 1,2 or 3. Every fractional class has an integral representative after clearing a denominator; the construction produces [A]⁻¹. To target a specified class, begin with an integral representative of its inverse. Thus all classes, not just the selected one, have such a small representative.

1 point: Derives ideal covolume and proves H is integral with the stated norm.
1 point: Obtains the exact integer cutoff.
1 point: Explains denominator clearing and coverage of every class through inversion.

(c)

A unital quotient of prime size ℓ is Fℓ. The image of s must satisfy X²+6=0. Over both F₂ and F₃ its only root is 0, so the quotient kernels are uniquely I=(2,s) and J=(3,s), respectively. Index 1 gives only O. Question 8 proves [I]=[J] is nonidentity of order two; hence all classes represented by these three possible ideals reduce to exactly [O] and [I]. The entire group is C₂. Product identities and norm obstructions had only proved the existence of a nontrivial order-two subgroup. The small-representative bound and unique small-index kernels supply the additional exhaustive step.

1 point: Uses the quotient-field/root argument to exhaust both small indices.
1 point: Combines the proved class identities and nonprincipality into the full group.
1 point: Identifies the precise extra role of the exhaustive bound.

Targeted lesson returns:

Turn the review into a new argument

Award points for reasoning actually present. Allow valid later reasoning based on a carried-forward value unless its criterion requires the correct value; do not award correct-result points for an incorrect result. Record missing hypotheses, unjustified inverses, invalid identifications and arithmetic errors separately.

The total is out of 100, with no university-grade conversion. Preserve your first attempt, consult the targeted lesson, explain the repair and attempt a fresh case. A teacher or peer may review the complete argument; if nobody did, label it self-review. Public worked solutions cannot certify independent proficiency.

Return to Abstract algebra II: rings and fields for more focused practice. For sustained synthesis, build a presentation-and-operator manuscript with exact certificates, revisions and a separate changed case.

Open the other abstract-algebra practice form · Return to the mathematics program