Polynomial Rings and Integral Domains Questions and Answers

Q: Let \(p(x) = 9x^{5} + 10x^{3} + 5x + 15\) and \(q(x) = x^{3} - x^{2} - x - 2\) be two polynomials in \(\mathbb{Q}[x]\). Then, over \(\mathbb{Q}\),

\(p(x)\) 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.

Q: Consider the polynomial \(p(X) = X^{4} + 4\) in the ring \(\mathbb{Q}[X]\) of polynomials in the variable \(X\) with coefficients in the field \(\mathbb{Q}\) of rational numbers. Then

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.

Q: For an odd prime \(p\), consider the ring \(\mathbb{Z}[\sqrt{-p}] = \{a + b\sqrt{-p} : a, b \in \mathbb{Z}\}\) \(\subseteq \mathbb{C}\). Then the element 2 in \(\mathbb{Z}[\sqrt{-p}]\) is

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.

Q: Let \(\mathbb{R}[X]\) denote the ring of polynomials in X with real coefficients. Then, the quotient ring \(\mathbb{R}[X]/(X^{4} + 4)\) is

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.

Q: Let \(\mathbb{R}[X^{2}, X^{3}]\) be the subring of \(\mathbb{R}[X]\) generated by \(X^{2}\) and \(X^{3}\). Consider the following statements.I. The ring \(\mathbb{R}[X^{2}, X^{3}]\) is a unique factorization domain.II. The ring \(\mathbb{R}[X^{2}, X^{3}]\) is a principal ideal domain.Which one of the following is correct?

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.

Q: Which of the following rings is a Unique Factorization domain but not a PID ?

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).

Q: Which of the following statements is false for \(\mathbf{Q}[\sqrt{2}]\) and \(\mathbf{Z}[\sqrt{2}]\) ?

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).

Q: Which of the following is a PID ?

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.

Q: Let \(n\) be a square free natural number. Then the quotient ring \(Q[X]/\lt X^{2} - n\gt\) is :

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.

Q: Let \(p\) be an odd prime number and \(\mathbf{Z}\left[\sqrt{-p}\right] = \{a + b\sqrt{-p} \mid a, b \in \mathbf{Z}\}\) \((\subseteq \mathbf{C})\). Then the ring \(\mathbf{Z}\left[\sqrt{-p}\right]\) is :

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.

Let \(p(x) = 9x^{5} + 10x^{3} + 5x + 15\) and \(q(x) = x^{3} - x^{2} - x - 2\) be two polynomials in \(\mathbb{Q}[x]\). Then, over \(\mathbb{Q}\),

A.\(p(x)\) and \(q(x)\) are both irreducible
B.\(p(x)\) is reducible but \(q(x)\) is irreducible
C.\(p(x)\) is irreducible but \(q(x)\) is reducible
D.\(p(x)\) and \(q(x)\) are both reducible

Exam Year(s)

GATE Mathematics - 2015,

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.

Discuss in Forum

Consider the polynomial \(p(X) = X^{4} + 4\) in the ring \(\mathbb{Q}[X]\) of polynomials in the variable \(X\) with coefficients in the field \(\mathbb{Q}\) of rational numbers. Then

A.the set of zeros of \(p(X)\) in \(\mathbb{C}\) forms a group under multiplication
B.\(p(X)\) is reducible in the ring \(\mathbb{Q}[X]\)
C.the splitting field of \(p(X)\) has degree 3 over \(\mathbb{Q}\)
D.the splitting field of \(p(X)\) has degree 4 over \(\mathbb{Q}\)

Exam Year(s)

GATE Mathematics - 2018,

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.

Discuss in Forum

For an odd prime \(p\), consider the ring \(\mathbb{Z}[\sqrt{-p}] = \{a + b\sqrt{-p} : a, b \in \mathbb{Z}\}\) \(\subseteq \mathbb{C}\). Then the element 2 in \(\mathbb{Z}[\sqrt{-p}]\) is

A.a unit
B.a square
C.a prime
D.irreducible

Exam Year(s)

GATE Mathematics - 2018,

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.

Discuss in Forum

Let \(\mathbb{R}[X]\) denote the ring of polynomials in X with real coefficients. Then, the quotient ring \(\mathbb{R}[X]/(X^{4} + 4)\) is

A.a field
B.an integral domain, but not a field
C.not an integral domain, but has 0 as the only nilpotent element
D.a ring which contains non-zero nilpotent elements

Exam Year(s)

GATE Mathematics - 2022,

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.

Discuss in Forum

Let \(\mathbb{R}[X^{2}, X^{3}]\) be the subring of \(\mathbb{R}[X]\) generated by \(X^{2}\) and \(X^{3}\). Consider the following statements.

I. The ring \(\mathbb{R}[X^{2}, X^{3}]\) is a unique factorization domain.
II. The ring \(\mathbb{R}[X^{2}, X^{3}]\) is a principal ideal domain.

Which one of the following is correct?

A.Both I and II are TRUE
B.I is TRUE and II is FALSE
C.I is FALSE and II is TRUE
D.Both I and II are FALSE

Exam Year(s)

GATE Mathematics - 2024,

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.

Discuss in Forum

Which of the following rings is a Unique Factorization domain but not a PID ?

A.\(\mathbf{Z}\)
B.F[\(x\)] (F a field)
C.\(\mathbf{Z}[x]\)
D.\(\mathbf{Z}[i]\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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).

Discuss in Forum

Which of the following statements is false for \(\mathbf{Q}[\sqrt{2}]\) and \(\mathbf{Z}[\sqrt{2}]\) ?

A.There are infinitely many units in \(\mathbf{Z}[\sqrt{2}]\)
B.\(\mathbf{Q}[\sqrt{2}]\) is not a subfield of complex numbers
C.\(\mathbf{Q}[\sqrt{2}]\) is isomorphic to \(\mathbf{Q}[x]/(x^{2} - 2)\)
D.There are infinitely many primes in \(\mathbf{Z}[\sqrt{2}]\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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).

Discuss in Forum

Which of the following is a PID ?

A.\(\mathbf{Z}[x]\)
B.\(\mathbf{R}(x)\ [y]\)
C.\(\mathbf{C}[x, y]\)
D.\(\mathbf{Z}_{30}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

Let \(n\) be a square free natural number. Then the quotient ring \(Q[X]/\lt X^{2} - n\gt\) is :

A.not a field
B.an integral domain which is not a field
C.not an integral domain
D.an integral domain

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

10.

Let \(p\) be an odd prime number and \(\mathbf{Z}\left[\sqrt{-p}\right] = \{a + b\sqrt{-p} \mid a, b \in \mathbf{Z}\}\) \((\subseteq \mathbf{C})\). Then the ring \(\mathbf{Z}\left[\sqrt{-p}\right]\) is :

A.not a Euclidean domain, but is a PID (Principal ideal domain)
B.Not a PID, but is an UFD
C.Neither Euclidean domain, nor a PID, but is an UFD
D.not an UFD

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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