(a) Consider the polynomial ring Z[x] over the ring Z of integers. Let S be an ideal of Z[x] generated by x. Show that S is prime but not a maximal ideal of Z[x]. (15 marks)

(b) Find the upper and lower Riemann integrals for the function f defined o

Question

(a) Consider the polynomial ring Z[x] over the ring Z of integers. Let S be an ideal of Z[x] generated by x. Show that S is prime but not a maximal ideal of Z[x]. (15 marks)

(b) Find the upper and lower Riemann integrals for the function f defined on [0, 1] as follows: f(x) = (1-x²)^½, if x is rational; f(x) = (1-x), if x is irrational. Hence, show that f is not Riemann integrable on [0, 1]. (15 marks)

(c) The personnel manager of a company wants to assign officers A, B and C to the regional offices at Delhi, Mumbai, Kolkata and Chennai. The cost of relocation (in thousand Rupees) of the three officers at the four regional offices are given below:

| Officer | Delhi | Mumbai | Kolkata | Chennai |
|---------|-------|--------|---------|---------|
| A | 16 | 22 | 24 | 20 |
| B | 10 | 32 | 26 | 16 |
| C | 10 | 20 | 46 | 30 |

Find the assignment which minimizes the total cost of relocation and also determine the minimum cost. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Let S = (x) in Z[x]. Define φ: Z[x] → Z by φ(p(x)) = p(0). This is a surjective ring homomorphism. Its kernel is exactly the set of polynomials with constant term 0, which is S. Hence Z[x]/S ≅ Z. Since Z is an integral domain, S is prime.

To see that S is not maximal, consider T = (2, x). Every element of T has even constant term, so 1 ∉ T; hence T is a proper ideal of Z[x]. Also S ⊊ T, because 2 ∈ T but 2 ∉ S. Thus S is contained properly in a proper ideal, so S is not maximal. Final: S is prime but not maximal.

(b) Let P = {0 = x₀ < x₁ < ... < xₙ = 1} be any partition, and let Iᵢ = [xᵢ₋₁, xᵢ]. Since rationals and irrationals are both dense in R, on each Iᵢ the rational values of f are √(1-x²) and the irrational values are 1-x. Both functions are decreasing on [0, 1].

For 0 ≤ t ≤ 1, √(1-t²) ≥ 1-t, because
1-t² - (1-t)² = 2t(1-t) ≥ 0.
Therefore, on Iᵢ,
Mᵢ = sup f = √(1-xᵢ₋₁²),
mᵢ = inf f = 1-xᵢ.
Hence the upper and lower Darboux sums are
U(P,f) = Σ Δxᵢ √(1-xᵢ₋₁²),
L(P,f) = Σ Δxᵢ (1-xᵢ),
where Δxᵢ = xᵢ - xᵢ₋₁.

The upper Riemann integral is
inf_P U(P,f) = ∫₀¹ √(1-x²) dx = π/4,
the area of a quarter unit circle. The lower Riemann integral is
sup_P L(P,f) = ∫₀¹ (1-x) dx = 1/2.

Since π/4 ≠ 1/2, the upper and lower integrals are unequal. Therefore f is not Riemann integrable on [0, 1]. Final: upper integral = π/4, lower integral = 1/2; f is not Riemann integrable.

(c) This is an assignment problem. Since there are 3 officers and 4 offices, add a dummy officer D with zero cost to every office to make a 4×4 matrix. The cost matrix in thousand Rupees is:
A: 16 22 24 20
B: 10 32 26 16
C: 10 20 46 30
D: 0 0 0 0

Use the Hungarian method. Row reduction gives:
A: 0 6 8 4
B: 0 22 16 6
C: 0 10 36 20
D: 0 0 0 0
Column minima are all 0, so the matrix is unchanged.

Cover zeros with the Delhi column and the dummy row D. The minimum uncovered entry is 4. Subtract it from uncovered entries and add it at intersections. This gives:
A: 0 2 4 0
B: 0 18 12 2
C: 0 6 32 16
D: 4 0 0 0

Next cover zeros with the Delhi column, the Chennai column, and the dummy row D. The minimum uncovered entry is 2. Subtract and add appropriately:
A: 0 0 2 0
B: 0 16 10 2
C: 0 4 30 16
D: 6 0 0 2

Next cover zeros with the Delhi column and rows A and D. The minimum uncovered entry is 2. Subtract and add appropriately:
A: 2 0 2 0
B: 0 14 8 0
C: 0 2 28 14
D: 8 0 0 2

Now an optimal zero assignment is:
A → Mumbai, B → Chennai, C → Delhi, D → Kolkata.

Thus the real assignment is A to Mumbai, B to Chennai, C to Delhi, leaving Kolkata vacant. The minimum total cost is
22 + 16 + 10 = 48 thousand Rupees = ₹48,000.

Final: A → Mumbai, B → Chennai, C → Delhi; minimum relocation cost = ₹48,000.

(a) Consider the polynomial ring Z[x] over the ring Z of integers. Let S be an ideal of Z[x] generated by x. Show that S is prime…

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