Fields and Galois Theory Questions and Answers

Q: Let K be an algebraically closed field containing a finite field F. Let L be the subfield of K consisting of elements of K that are algebraic over F.Consider the following statements:S1: L is algebraically closed.S2: L is infinite.Then, which one of the following is correct?

Both S1 and S2 are TRUE.L, the set of elements of K algebraic over F, is the algebraic closure of F inside K. It is algebraically closed: a root in K of a polynomial with coefficients in L is algebraic over L, hence over F, so it lies in L; and K supplies a root because it is algebraically closed.An algebraically closed field is always infinite: if a field were finite with elements \(a_{1}, \ldots, a_{n}\), the polynomial \((x - a_{1})\cdots(x - a_{n}) + 1\) would have no root in it. Concretely, L contains the fields \(\mathbb{F}_{p^{n}}\) for every n. So both statements hold.

Q: Let m be an even positive integer. What is the product of all mth roots of unity in C ?

For even m the product is -1.The mth roots of unity are the roots of \(z^{m} - 1 = 0\). For a monic polynomial of degree m, the product of the roots is \((-1)^{m}\) times the constant term, here \((-1)^{m}(-1) = (-1)^{m + 1}\). When m is even this is \(-1\).A direct check: pair each root \(\zeta\) with its conjugate \(\bar{\zeta} = \zeta^{-1}\); each pair multiplies to 1. The only roots equal to their own inverses are 1 and \(-1\), and \(-1\) is a root exactly when m is even. So the product is \(1 \cdot (-1) = -1\). For m = 2, the roots 1 and \(-1\) give \(-1\); for m = 4, \(1 \cdot i \cdot (-1) \cdot (-i) = -1\).1 is the answer for odd m, and \(\pm i\) cannot occur because the non-real roots come in conjugate pairs.

Q: Let F be a field with 128 elements, then which of the following is true ?

The true statement is F has no proper non-prime subfield.\(128 = 2^{7}\), so F is \(\mathbf{F}_{2^{7}}\). The subfields of \(\mathbf{F}_{p^{n}}\) are exactly the fields \(\mathbf{F}_{p^{d}}\) with d dividing n. Since 7 is prime, the only divisors are 1 and 7: the subfields are the prime field \(\mathbf{F}_{2}\) and F itself. So F has no proper subfield other than the prime field.A subfield with 8 elements would need \(3 \mid 7\), and one with 32 elements would need \(5 \mid 7\); neither holds. And a field with 128 elements does exist, since a finite field of order \(p^{n}\) exists for every prime p and every \(n \ge 1\).

Q: Let F be a field with at least two elements. Then which of the following statements need not be true ?

The statement that need not be true is If F is infinite, then \(\mathbf{Z}_{p}\) cannot be embedded in F for any prime \(p\).The field of rational functions \(\mathbf{Z}_{p}(x)\), or the algebraic closure of \(\mathbf{Z}_{p}\), is infinite but has characteristic \(p\) and contains \(\mathbf{Z}_{p}\).The other three always hold. Every field contains its prime subring, which is \(\mathbf{Z}\) (characteristic 0) or \(\mathbf{Z}_{p}\) (characteristic \(p\)), giving A. A field has no zero divisors, and \(\mathbf{Z}_{n}\) for composite \(n\) has them, so it cannot sit inside F (B). And if \(\mathbf{Z}\) embeds, every nonzero integer is invertible in F, so all fractions \(a/b\) are there and Q embeds (C).

Q: The degree of the splitting field of the polynomial \(x^{3} - 2 \in \mathbf{Q}[x]\) is :

The degree is 6.The roots of \(x^{3} - 2\) are \(\sqrt[3]{2}\), \(\sqrt[3]{2}\,\omega\) and \(\sqrt[3]{2}\,\omega^{2}\), where \(\omega = e^{2\pi i/3}\). The splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).\(x^{3} - 2\) is irreducible by Eisenstein at 2, so \([\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] = 3\). That field is real, so it does not contain \(\omega\), which satisfies \(x^{2} + x + 1\); adjoining \(\omega\) doubles the degree:\[ \begin{aligned} [\mathbf{Q}(\sqrt[3]{2}, \omega) : \mathbf{Q}] \\ &= 3 \times 2 = 6 \end{aligned} \]3 is the degree of one real root only. The Galois group is \(S_{3}\), of order 6, which also rules out 4 and 5.

Q: Which of the following constructions is not possible by ruler and compass ?

The construction that is not possible is a regular polygon of 7 sides.By Gauss and Wantzel, a regular \(n\)-gon is constructible with ruler and compass exactly when \(n\) is a power of 2 times a product of distinct Fermat primes (3, 5, 17, 257, 65537). 7 is prime but not a Fermat prime; equivalently \(\cos(2\pi/7)\) has degree 3 over Q, not a power of 2.The others are possible. Trisecting 90° means constructing 30°, which is easy. \(1 + \sqrt{2} + \sqrt[4]{2}\) uses only repeated square roots. And \(e^{\pi i/40}\) is a vertex of the regular 80-gon, and \(80 = 2^{4} \times 5\) with 5 a Fermat prime, so it is constructible.

Q: Let \(p\) be a prime number and \(\mathbf{F}_{p}\) be a finite field with \(p\) elements. Then the unit group \(\mathbf{F}_{p}^{x} = \mathbf{F}_{p} \backslash \{0\}\) of \(\mathbf{F}_{p}\) is simple if and only if :

The unit group is simple if and only if \(p = 3\).\(\mathbf{F}_{p}^{\times}\) is cyclic of order \(p - 1\), and a cyclic group is simple exactly when its order is prime. So we need \(p - 1\) prime. Since \(p - 1\) is even for every odd prime, it can be prime only if \(p - 1 = 2\), that is \(p = 3\).For \(p = 2\) the unit group is trivial, and the trivial group is not counted as simple. For \(p \ge 5\), \(p - 1\) is an even number at least 4, so the group has a proper non-trivial subgroup. Fermat primes \(2^{n} + 1\) with \(n \ge 2\) give \(p - 1 = 2^{n}\), again composite.

Q: The quadratic form associated to the trace form \(\mathrm{Tr}_{\mathbf R}^{\mathbf C} : \mathbf C \times \mathbf C \to \mathbf R\), \((x, y) \mapsto \mathrm{Tr}(\lambda_{XY})\), where \(\lambda_{XY} : \mathbf{C} \to \mathbf{C}\) is the \(\mathbf{R}\)-linear map defined by \(\lambda_{XY}(z) = xyz\), is :

The quadratic form is \(X^{2} - Y^{2}\).For \(w \in \mathbf{C}\), multiplication by \(w = a + ib\) is the \(\mathbf{R}\)-linear map with matrix \(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\) on the basis 1, \(i\), so its trace is \(2a = 2\,\mathrm{Re}\,w\). Hence the trace form is \(B(x, y) = 2\,\mathrm{Re}(xy)\).Its quadratic form at \(z = X + iY\) uses \(z^{2} = X^{2} - Y^{2} + 2iXY\):\[ B(z, z) = 2\,\mathrm{Re}(z^{2}) = 2(X^{2} - Y^{2}) \]With the usual convention \(q(z) = \tfrac{1}{2}B(z, z)\), this is \(X^{2} - Y^{2}\); with \(q = B(z, z)\) it differs only by the factor 2. Either way it is indefinite, of signature (1, 1), which rules out \(X^{2}\), \(Y^{2}\) and the positive definite \(X^{2} + Y^{2}\).

Q: Let \(p\) be a prime number and \(n\), \(m\) natural numbers with \(m\) divides \(n\). Then the finite extension \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) of finite fields is a :

It is a Galois extension with cyclic Galois group of order \(n/m\).The printed \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) is the extension of the field with \(p^{m}\) elements by the field with \(p^{n}\) elements. Every finite extension of finite fields is Galois (the larger field is the splitting field of \(x^{p^{n}} - x\)), and its degree is \(n/m\).The Galois group is generated by the Frobenius map \(a \mapsto a^{p^{m}}\), so it is cyclic of order equal to the degree, \(n/m\).Non-cyclic groups are wrong, since the group is generated by one element, and order \(nm\) confuses the product with the quotient of the degrees.

Q: Let K/F be a Galois extension of fields. If Gal(K/F) is isomoprhic to \(S_{3}\), the number of fields L such that \(F \subseteq L \subseteq K\) is :

There are 6 intermediate fields.By the fundamental theorem of Galois theory, the fields L with \(F \subseteq L \subseteq K\) correspond one-to-one with the subgroups of Gal(K/F). \(S_{3}\) has exactly six subgroups:the trivial subgroup (giving L = K)three subgroups of order 2, generated by the transpositionsthe alternating group \(A_{3}\) of order 3\(S_{3}\) itself (giving L = F)So there are 6 fields, counting F and K themselves. 4 forgets the two trivial ends, and 8 or 10 overcount the subgroups.

Let K be an algebraically closed field containing a finite field F. Let L be the subfield of K consisting of elements of K that are algebraic over F.

Consider the following statements:

S1: L is algebraically closed.
S2: L is infinite.

Then, which one of the following is correct?

A.S1 is TRUE and S2 is FALSE
B.S2 is TRUE and S1 is FALSE
C.both S1 and S2 are TRUE
D.neither S1 nor S2 is TRUE

Exam Year(s)

GATE Mathematics - 2025,

Answer: C

Both S1 and S2 are TRUE.

L, the set of elements of K algebraic over F, is the algebraic closure of F inside K. It is algebraically closed: a root in K of a polynomial with coefficients in L is algebraic over L, hence over F, so it lies in L; and K supplies a root because it is algebraically closed.

An algebraically closed field is always infinite: if a field were finite with elements \(a_{1}, \ldots, a_{n}\), the polynomial \((x - a_{1})\cdots(x - a_{n}) + 1\) would have no root in it. Concretely, L contains the fields \(\mathbb{F}_{p^{n}}\) for every n. So both statements hold.

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Let m be an even positive integer. What is the product of all mth roots of unity in C ?

A.1
B.-1
C.\(i\)
D.\(-i\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

For even m the product is -1.

The mth roots of unity are the roots of \(z^{m} - 1 = 0\). For a monic polynomial of degree m, the product of the roots is \((-1)^{m}\) times the constant term, here \((-1)^{m}(-1) = (-1)^{m + 1}\). When m is even this is \(-1\).

A direct check: pair each root \(\zeta\) with its conjugate \(\bar{\zeta} = \zeta^{-1}\); each pair multiplies to 1. The only roots equal to their own inverses are 1 and \(-1\), and \(-1\) is a root exactly when m is even. So the product is \(1 \cdot (-1) = -1\). For m = 2, the roots 1 and \(-1\) give \(-1\); for m = 4, \(1 \cdot i \cdot (-1) \cdot (-i) = -1\).

1 is the answer for odd m, and \(\pm i\) cannot occur because the non-real roots come in conjugate pairs.

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Let F be a field with 128 elements, then which of the following is true ?

A.F has a subfield with 8 elements
B.F has no proper non-prime subfield
C.Such a field F does not exist
D.F has a subfield with 32 elements

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The true statement is F has no proper non-prime subfield.

\(128 = 2^{7}\), so F is \(\mathbf{F}_{2^{7}}\). The subfields of \(\mathbf{F}_{p^{n}}\) are exactly the fields \(\mathbf{F}_{p^{d}}\) with d dividing n. Since 7 is prime, the only divisors are 1 and 7: the subfields are the prime field \(\mathbf{F}_{2}\) and F itself. So F has no proper subfield other than the prime field.

A subfield with 8 elements would need \(3 \mid 7\), and one with 32 elements would need \(5 \mid 7\); neither holds. And a field with 128 elements does exist, since a finite field of order \(p^{n}\) exists for every prime p and every \(n \ge 1\).

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Let F be a field with at least two elements. Then which of the following statements need not be true ?

A.Either \(\mathbf{Z}\) or \(\mathbf{Z}_{p}\) for some integer \(p \ge 2\) is embedded in F
B.If \(n\) is not prime, then \(\mathbf{Z}_{n}\) is not embedded in F
C.If \(\mathbf{Z}\) is embedded in F, then the field Q is also embedded in F
D.If F is infinite, then \(\mathbf{Z}_{p}\) cannot be embedded in F for any prime \(p\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The statement that need not be true is If F is infinite, then \(\mathbf{Z}_{p}\) cannot be embedded in F for any prime \(p\).

The field of rational functions \(\mathbf{Z}_{p}(x)\), or the algebraic closure of \(\mathbf{Z}_{p}\), is infinite but has characteristic \(p\) and contains \(\mathbf{Z}_{p}\).

The other three always hold. Every field contains its prime subring, which is \(\mathbf{Z}\) (characteristic 0) or \(\mathbf{Z}_{p}\) (characteristic \(p\)), giving A. A field has no zero divisors, and \(\mathbf{Z}_{n}\) for composite \(n\) has them, so it cannot sit inside F (B). And if \(\mathbf{Z}\) embeds, every nonzero integer is invertible in F, so all fractions \(a/b\) are there and Q embeds (C).

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The degree of the splitting field of the polynomial \(x^{3} - 2 \in \mathbf{Q}[x]\) is :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The degree is 6.

The roots of \(x^{3} - 2\) are \(\sqrt[3]{2}\), \(\sqrt[3]{2}\,\omega\) and \(\sqrt[3]{2}\,\omega^{2}\), where \(\omega = e^{2\pi i/3}\). The splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).

\(x^{3} - 2\) is irreducible by Eisenstein at 2, so \([\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] = 3\). That field is real, so it does not contain \(\omega\), which satisfies \(x^{2} + x + 1\); adjoining \(\omega\) doubles the degree:

\[ \begin{aligned} [\mathbf{Q}(\sqrt[3]{2}, \omega) : \mathbf{Q}] \\ &= 3 \times 2 = 6 \end{aligned} \]

3 is the degree of one real root only. The Galois group is \(S_{3}\), of order 6, which also rules out 4 and 5.

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Which of the following constructions is not possible by ruler and compass ?

A.Trisection of an angle of 90°
B.Construction of a regular polygon of 7 sides
C.Construction of length equal to \(1 + \sqrt{2} + \sqrt[4]{2}\)
D.Construction of the complex number \(e^{\pi i/40}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The construction that is not possible is a regular polygon of 7 sides.

By Gauss and Wantzel, a regular \(n\)-gon is constructible with ruler and compass exactly when \(n\) is a power of 2 times a product of distinct Fermat primes (3, 5, 17, 257, 65537). 7 is prime but not a Fermat prime; equivalently \(\cos(2\pi/7)\) has degree 3 over Q, not a power of 2.

The others are possible. Trisecting 90° means constructing 30°, which is easy. \(1 + \sqrt{2} + \sqrt[4]{2}\) uses only repeated square roots. And \(e^{\pi i/40}\) is a vertex of the regular 80-gon, and \(80 = 2^{4} \times 5\) with 5 a Fermat prime, so it is constructible.

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Let \(p\) be a prime number and \(\mathbf{F}_{p}\) be a finite field with \(p\) elements. Then the unit group \(\mathbf{F}_{p}^{x} = \mathbf{F}_{p} \backslash \{0\}\) of \(\mathbf{F}_{p}\) is simple if and only if :

A.\(p = 2\)
B.\(p = 3\)
C.\(p \ge 5\)
D.\(p\) is of the form \(2^{n} + 1\) for some \(n \ge 2\).

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The unit group is simple if and only if \(p = 3\).

\(\mathbf{F}_{p}^{\times}\) is cyclic of order \(p - 1\), and a cyclic group is simple exactly when its order is prime. So we need \(p - 1\) prime. Since \(p - 1\) is even for every odd prime, it can be prime only if \(p - 1 = 2\), that is \(p = 3\).

For \(p = 2\) the unit group is trivial, and the trivial group is not counted as simple. For \(p \ge 5\), \(p - 1\) is an even number at least 4, so the group has a proper non-trivial subgroup. Fermat primes \(2^{n} + 1\) with \(n \ge 2\) give \(p - 1 = 2^{n}\), again composite.

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The quadratic form associated to the trace form \(\mathrm{Tr}_{\mathbf R}^{\mathbf C} : \mathbf C \times \mathbf C \to \mathbf R\), \((x, y) \mapsto \mathrm{Tr}(\lambda_{XY})\), where \(\lambda_{XY} : \mathbf{C} \to \mathbf{C}\) is the \(\mathbf{R}\)-linear map defined by \(\lambda_{XY}(z) = xyz\), is :

A.\(X^{2}\)
B.\(Y^{2}\)
C.\(X^{2} + Y^{2}\)
D.\(X^{2} - Y^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The quadratic form is \(X^{2} - Y^{2}\).

For \(w \in \mathbf{C}\), multiplication by \(w = a + ib\) is the \(\mathbf{R}\)-linear map with matrix \(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\) on the basis 1, \(i\), so its trace is \(2a = 2\,\mathrm{Re}\,w\). Hence the trace form is \(B(x, y) = 2\,\mathrm{Re}(xy)\).

Its quadratic form at \(z = X + iY\) uses \(z^{2} = X^{2} - Y^{2} + 2iXY\):

\[ B(z, z) = 2\,\mathrm{Re}(z^{2}) = 2(X^{2} - Y^{2}) \]

With the usual convention \(q(z) = \tfrac{1}{2}B(z, z)\), this is \(X^{2} - Y^{2}\); with \(q = B(z, z)\) it differs only by the factor 2. Either way it is indefinite, of signature (1, 1), which rules out \(X^{2}\), \(Y^{2}\) and the positive definite \(X^{2} + Y^{2}\).

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Let \(p\) be a prime number and \(n\), \(m\) natural numbers with \(m\) divides \(n\). Then the finite extension \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) of finite fields is a :

A.Galois extension with abelian but non-cyclic Galois group of order \(\dfrac{n}{m}\)
B.Galois extension with cyclic Galois group of order \(\dfrac{n}{m}\)
C.Galois extension with abelian but non-cyclic Galois group of order \(nm\)
D.Galois extension with cyclic Galois group of order \(nm\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

It is a Galois extension with cyclic Galois group of order \(n/m\).

The printed \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) is the extension of the field with \(p^{m}\) elements by the field with \(p^{n}\) elements. Every finite extension of finite fields is Galois (the larger field is the splitting field of \(x^{p^{n}} - x\)), and its degree is \(n/m\).

The Galois group is generated by the Frobenius map \(a \mapsto a^{p^{m}}\), so it is cyclic of order equal to the degree, \(n/m\).

Non-cyclic groups are wrong, since the group is generated by one element, and order \(nm\) confuses the product with the quotient of the degrees.

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10.

Let K/F be a Galois extension of fields. If Gal(K/F) is isomoprhic to \(S_{3}\), the number of fields L such that \(F \subseteq L \subseteq K\) is :

A.6
B.4
C.10
D.8

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

There are 6 intermediate fields.

By the fundamental theorem of Galois theory, the fields L with \(F \subseteq L \subseteq K\) correspond one-to-one with the subgroups of Gal(K/F). \(S_{3}\) has exactly six subgroups:

the trivial subgroup (giving L = K)
three subgroups of order 2, generated by the transpositions
the alternating group \(A_{3}\) of order 3
\(S_{3}\) itself (giving L = F)

So there are 6 fields, counting F and K themselves. 4 forgets the two trivial ends, and 8 or 10 overcount the subgroups.

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