Polynomial Rings and Integral Domains Questions and Answers

Answer: C

The integral domain that is not a field is \(\mathbf{R}[x]\).

The polynomial ring over a field has no zero divisors, since the leading terms of two non-zero polynomials multiply to a non-zero term, so it is an integral domain. But \(x\) has no inverse: \(x \cdot p(x)\) always has degree at least 1, never equal to 1. So it is not a field.

\(\mathbf{Z}_{10}\) is not even an integral domain (\(2 \cdot 5 = 0\)). \(\mathbf{Z}_{31}\) is a field because 31 is prime, and \(\mathbf{C}(x)\), the field of rational functions, is a field by construction.

Answer: A

\(X^{16} + 1\) has 0 linear factors in R[X].

A linear factor \(X - c\) with \(c\) real would mean \(c^{16} + 1 = 0\). But \(c^{16} \ge 0\) for every real \(c\), so \(c^{16} + 1 \ge 1 \gt 0\). The polynomial has no real roots and hence no real linear factors.

Its 16 complex roots, the 32nd roots of unity that are not 16th roots of unity, come in conjugate pairs, so over R it splits into 8 irreducible quadratics and nothing else. That rules out 1, 2 and 4.

Answer: A

\(X^{8} + 1\) has 4 irreducible quadratic factors in R[X].

Its roots are the eight complex numbers \(e^{i\pi(2k + 1)/8}\), \(k = 0, \ldots, 7\). None is real, since \(x^{8} + 1 \gt 0\) for real \(x\), and they come in four conjugate pairs. Each pair gives a real irreducible quadratic \(X^{2} - 2\cos\theta\, X + 1\), so

\[ \begin{aligned} X^{8} + 1 = \prod_{k=0}^{3} \Big(&X^{2} \\ &- 2\cos\tfrac{(2k+1)\pi}{8}X + 1\Big) \end{aligned} \]

Four quadratics account for all eight degrees, so 3, 2 or 1 quadratic factors would leave real roots, which do not exist.

Answer: A

There are 5 such polynomials.

Count by degree over \(\mathbf{Z}_{2}\):

• Degree 1: \(x\) and \(x + 1\), both irreducible. That is 2.
• Degree 2: a quadratic is irreducible exactly when it has no root in \(\mathbf{Z}_{2}\). Of \(x^{2}, x^{2} + 1, x^{2} + x, x^{2} + x + 1\), only \(x^{2} + x + 1\) has no root (\(x^{2} + 1 = (x + 1)^{2}\)). That is 1.
• Degree 3: again irreducible exactly when there is no root. The constant term must be 1 (else 0 is a root) and the number of terms must be odd (else 1 is a root), leaving \(x^{3} + x + 1\) and \(x^{3} + x^{2} + 1\). That is 2.

Total \(2 + 1 + 2 = 5\). The count 3 forgets the linear polynomials, which are always irreducible, and 4 or 6 come from miscounting the cubics.

Answer: B

The false statement is Every principal ideal domain is a Euclidean domain.

Every Euclidean domain is a PID, but not conversely. The standard counterexample is \(\mathbf{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]\), which is a PID but admits no Euclidean function.

The other three are true. K[x] over a field is Euclidean, with the degree as the Euclidean function (division with remainder). Since K[x] is both Euclidean and a PID, the biconditional 'Euclidean if and only if PID' holds for it trivially, both sides being true. And K[x, y] is not a PID: the ideal \((x, y)\) needs two generators, since a single generator would have to divide both \(x\) and \(y\) and so be a unit.

Answer: D

Neither (I) nor (II) is true.

Z[x] is not a principal ideal domain. The ideal \((2, x)\), the polynomials with even constant term, is not principal: a generator would divide both 2 and \(x\), so it would be \(\pm 1\), but \(1 \notin (2, x)\). Another way to see it: \(\mathbf{Z}[x]/(x) \cong \mathbf{Z}\) is not a field, so \((x)\) is prime but not maximal, which cannot happen for a nonzero prime ideal in a PID.

Since every Euclidean domain is a PID, Z[x] is not Euclidean either, so (II) fails as well. What is true is that Z[x] is a unique factorisation domain (Gauss's lemma), which is the property candidates often confuse with being a PID.

Answer: D

The reducible polynomial is x⁹ + 5x⁵ + x⁴ + 5.

Group the terms:

\[ \begin{aligned} x^{9} + 5x^{5} + x^{4} + 5 &= x^{5}(x^{4} + 5) + (x^{4} + 5) \\ &= (x^{5} + 1)(x^{4} + 5) \end{aligned} \]

The other three are irreducible by Eisenstein's criterion: \(x^{11} - 11x - 11\) with \(p = 11\), \(x^{10} - 7\) with \(p = 7\), and \(x^{5} - 5\) with \(p = 5\). In each case p divides every coefficient except the leading one, and \(p^{2}\) does not divide the constant term.

Answer: B

Only (II) is true.

(I) In \(\mathbf{Z}[i\sqrt{5}]\),

\[ 3 \cdot 3 = 9 = (2 + i\sqrt{5})(2 - i\sqrt{5}) \]

3 divides the product but neither factor, since \((2 \pm i\sqrt{5})/3\) is not in the ring. So 3 is not prime there.

(II) In \(\mathbf{Z}/10\mathbf{Z}\) the ideal generated by \(\overline{2}\) is {0, 2, 4, 6, 8}, and the quotient by it is \(\mathbf{Z}/2\mathbf{Z}\), a field, so \(\overline{2}\) is prime. But \(\overline{2} = \overline{2} \cdot \overline{6}\) (since 12 leaves remainder 2), and neither \(\overline{2}\) nor \(\overline{6}\) is a unit, so \(\overline{2}\) is not irreducible. Statement (II) is therefore true.

Answer: B

\(x^{3} - 9\) is irreducible in \(\mathbf{Z}_{n}[x]\) for \(n = 31\).

A cubic over a field is irreducible exactly when it has no root, so the question is whether 9 is a cube mod n.

Mod 31: 3 is a primitive root (its order is 30, since \(3^{15} \equiv -1\) and \(3^{10} \equiv 25\)). Then \(9 = 3^{2}\), and an element \(3^{k}\) is a cube in a cyclic group of order 30 only when 3 divides k. As 3 does not divide 2, 9 is not a cube, so \(x^{3} - 9\) has no root and is irreducible.

Mod 11, the group of units has order 10, prime to 3, so cubing is a bijection and every element, 9 included, is a cube; the cubic has a root. In \(\mathbf{Z}_{9}\) the polynomial is \(x^{3}\), plainly reducible, and in \(\mathbf{Z}_{10}\) it is \(x^{3} + 1 = (x + 1)(x^{2} - x + 1)\).

Answer: D

Neither I nor II is true.

I fails: the field C is trivially a UFD, but its subring \(\mathbf{Z}[\sqrt{-5}]\) is not, since \(6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})\) are two different factorisations into irreducibles.

II fails: in the UFD \(\mathbf{Z}[x]\), \(\gcd(2, x) = 1\), but the ideal \((2) + (x)\) consists of polynomials with even constant term and does not contain 1. The equality \((d) = (a) + (b)\) holds in a principal ideal domain, not in every UFD.