Polynomial Rings and Integral Domains Questions and Answers

Answer: C

\(p(x)\) is irreducible but \(q(x)\) is reducible.

\(p(x) = 9x^{5} + 10x^{3} + 5x + 15\) satisfies Eisenstein's criterion at the prime 5: 5 divides 10, 0 (the other coefficients), 5 and 15; it does not divide the leading coefficient 9; and \(5^{2} = 25\) does not divide the constant 15. So \(p\) is irreducible over \(\mathbb{Q}\).

\(q(2) = 8 - 4 - 2 - 2 = 0\), so \(x - 2\) is a factor: \(q(x) = (x - 2)(x^{2} + x + 1)\), which is reducible.

That rules out the options calling \(q\) irreducible or \(p\) reducible.

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.

Answer: D

The element 2 is irreducible.

Use the norm \(N(a + b\sqrt{-p}) = a^{2} + pb^{2}\), which is multiplicative. \(N(2) = 4\). A factorization \(2 = \alpha\beta\) into non-units would need \(N(\alpha) = N(\beta) = 2\), that is \(a^{2} + pb^{2} = 2\), which has no integer solution when \(p \ge 3\). So 2 is irreducible.

It is not a unit, since units have norm 1. It is not a square: \((a + b\sqrt{-p})^{2} = 2\) forces \(ab = 0\) and then \(a^{2} = 2\) or \(-pb^{2} = 2\). It is not prime: \(p + 1\) is even, so 2 divides \((1 + \sqrt{-p})(1 - \sqrt{-p}) = 1 + p\), yet 2 divides neither factor, because \(\frac{1}{2} \pm \frac{1}{2}\sqrt{-p}\) is not in the ring.

Answer: C

The quotient ring is not an integral domain, but has 0 as the only nilpotent element.

\(X^{4} + 4\) factors over \(\mathbb{R}\) (Sophie Germain's identity):

\[ X^{4} + 4 = (X^{2} + 2X + 2)(X^{2} - 2X + 2) \]

Both quadratics have discriminant 4 - 8 = -4 \(\lt\) 0, so they are irreducible, and they are distinct, hence coprime. By the Chinese remainder theorem the quotient splits into two factors, each a copy of \(\mathbb{C}\):

\[ \begin{aligned} \mathbb{R}[X]/(X^{4}+4) &\cong \mathbb{R}[X]/(X^{2}+2X+2) \\ &\times \mathbb{R}[X]/(X^{2}-2X+2) \\ &\cong \mathbb{C} \times \mathbb{C} \end{aligned} \]

A product of two fields has zero divisors, such as (1, 0)(0, 1) = 0, so it is neither a field nor an integral domain. It has no non-zero nilpotents, because a nilpotent pair would need each coordinate nilpotent in a field, that is zero. Non-zero nilpotents would appear only if \(X^{4} + 4\) had a repeated factor.

Answer: D

Both I and II are FALSE.

In \(R = \mathbb{R}[X^{2}, X^{3}]\) there is no element of degree 1, so \(X^{2}\) and \(X^{3}\) are irreducible (any factorisation would need a factor of degree 1). Then \(X^{6} = X^{2}\cdot X^{2}\cdot X^{2} = X^{3}\cdot X^{3}\) gives two factorisations into irreducibles with different numbers of factors, so \(R\) is not a UFD.

Every principal ideal domain is a unique factorization domain, so \(R\) is not a PID either. That rules out every option in which I or II is true.

Answer: C

The UFD that is not a PID is \(\mathbf{Z}[x]\).

By Gauss's lemma, a polynomial ring over a UFD is a UFD, so \(\mathbf{Z}[x]\) has unique factorisation. But the ideal \((2, x)\), the polynomials with even constant term, is not principal: a generator would have to divide both 2 and \(x\), so it would be \(\pm 1\), yet \((2, x) \ne \mathbf{Z}[x]\) because 1 is not in it.

\(\mathbf{Z}\), \(F[x]\) over a field and the Gaussian integers \(\mathbf{Z}[i]\) are all Euclidean domains, hence PIDs (and so also UFDs).

Answer: B

The false statement is \(\mathbf{Q}[\sqrt{2}]\) is not a subfield of complex numbers.

\(\mathbf{Q}[\sqrt{2}] = \{a + b\sqrt{2} : a, b \in \mathbf{Q}\}\) is a set of real numbers closed under the field operations, with \((a + b\sqrt{2})^{-1} = (a - b\sqrt{2})/(a^{2} - 2b^{2})\). It is a subfield of \(\mathbf{R}\) and so of \(\mathbf{C}\).

The other statements are true. \(1 + \sqrt{2}\) is a unit of \(\mathbf{Z}[\sqrt{2}]\) (its inverse is \(\sqrt{2} - 1\)), and its powers are all distinct units (A). Since \(x^{2} - 2\) is irreducible over Q, evaluating at \(\sqrt{2}\) gives \(\mathbf{Q}[x]/(x^{2} - 2) \cong \mathbf{Q}[\sqrt{2}]\) (C). And \(\mathbf{Z}[\sqrt{2}]\) has infinitely many primes, for example a prime factor of each rational prime (D).

Answer: B

The PID is \(\mathbf{R}(x)[y]\).

\(\mathbf{R}(x)\) is the field of rational functions in \(x\), and a polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain.

The others are not. In \(\mathbf{Z}[x]\) the ideal \((2, x)\) is not principal. In \(\mathbf{C}[x, y]\) the ideal \((x, y)\) is not principal. And \(\mathbf{Z}_{30}\) is not an integral domain at all, since \(2 \cdot 15 = 0\), so it cannot be a PID even though all its ideals are principal.

Answer: D

The keyed answer is an integral domain.

For a square-free \(n \gt 1\), \(\sqrt{n}\) is irrational, so \(X^{2} - n\) has no rational root and, being of degree 2, is irreducible over \(Q\). The ideal it generates is then maximal, and \(Q[X]/\langle X^{2} - n \rangle \cong Q(\sqrt{n})\) is a field. Every field is an integral domain, so (D) holds, while 'not a field' and 'an integral domain which is not a field' are false.

The key assumes \(n \gt 1\): for \(n = 1\), \(X^{2} - 1 = (X - 1)(X + 1)\) and the quotient has zero divisors, which is the only case in which 'not an integral domain' would apply.

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.