Fields and Galois Theory Questions and Answers

Answer: A

\(\sigma\) is bijective.

A ring homomorphism between fields that sends 1 to 1 is injective, because its kernel is an ideal of a field not containing 1, hence zero. For surjectivity, take \(\alpha \in L\) with minimal polynomial \(p\) over K, and let \(R\) be the finite set of roots of \(p\) in L. Since \(\sigma\) fixes K, it maps \(R\) into \(R\), and being injective on a finite set it maps \(R\) onto \(R\). So \(\alpha = \sigma(\beta)\) for some \(\beta \in R\).

Injectivity rules out C and D. B can happen only for transcendental extensions, such as \(x \mapsto x^{2}\) on \(K(x)\), not for algebraic ones.

Answer: A

The field is not equal to \(\mathbf{Q}(\sqrt{30})\).

\(\mathbf{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\) has degree 8 over Q, while \(\mathbf{Q}(\sqrt{30})\) has degree 2. It is a proper subfield (it contains \(\sqrt{30} = \sqrt{2}\sqrt{3}\sqrt{5}\)) but cannot be the whole field.

The others equal it. A sum \(\sqrt{2} \pm \sqrt{3} \pm \sqrt{5}\) has eight distinct conjugates under the Galois group, so it is a primitive element (B, D). And \(\sqrt{2} + \sqrt{3}\) generates \(\mathbf{Q}(\sqrt{2}, \sqrt{3})\), so adjoining \(\sqrt{5}\) gives the whole field (C).

Answer: A

The splitting field of \(x^{3} - 2\) over Q has degree 6.

The roots are \(\sqrt[3]{2}\), \(\omega\sqrt[3]{2}\) and \(\omega^{2}\sqrt[3]{2}\), where \(\omega\) is a primitive cube root of unity, so the splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).

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

The first degree is 3 because \(x^{3} - 2\) is irreducible (Eisenstein at 2). The second is 2 because \(\omega\) satisfies \(x^{2} + x + 1\) and is not real, while \(\mathbf{Q}(\sqrt[3]{2})\) is real. The Galois group is \(S_{3}\), of order 6. The answer 3 stops at the real root, and 2 and 4 do not divide in the right way.

Answer: A

The values are α = 0 and β = −1.

For the sum: if q is odd and not 2, pick \(c \neq 0, 1\) in the field. Multiplying every element by c permutes the field, so \(c\alpha = \alpha\), hence \((c - 1)\alpha = 0\) and \(\alpha = 0\). (For a prime field this is just \(0 + 1 + \ldots + (p - 1) = p(p - 1)/2 \equiv 0\).)

For the product: each non-zero element pairs with its inverse, and the only elements equal to their own inverses are the roots of \(x^{2} = 1\), namely 1 and \(-1\), which are distinct when q is odd. All the pairs multiply to 1, leaving

\[ \beta = 1 \cdot (-1) = -1, \]

the generalised Wilson's theorem. The options with \(\alpha = 1\) or \(\beta = 1\) fail one of these two calculations.

Answer: C

F is the splitting field of \(x^{p^{n}} - x\) over \(\mathbf{F}_{p}\).

A finite field of characteristic p contains \(\mathbf{F}_{p}\) and is a vector space over it of some dimension n, so \(|F| = p^{n}\). Its multiplicative group has order \(p^{n} - 1\), so every non-zero element satisfies \(a^{p^{n} - 1} = 1\), and every element satisfies \(a^{p^{n}} = a\). So the \(p^{n}\) elements of F are exactly the roots of \(x^{p^{n}} - x\), and F is its splitting field.

F is not \(\mathbf{Z}_{p^{n}}\), which is not a field for \(n \gt 1\), nor the product ring \(\mathbf{F}_{p} \times \ldots \times \mathbf{F}_{p}\), which has zero divisors. The roots of \(x^{p^{n}} - 1\) are all equal to 1 in characteristic p, since \(x^{p^{n}} - 1 = (x - 1)^{p^{n}}\).

Answer: C

The minimal polynomial is \(x^{4} - 10x^{2} + 1\).

Let \(x = \sqrt{2} + \sqrt{3}\) and square twice:

\[ \begin{aligned} x^{2} &= 5 + 2\sqrt{6} \\ (x^{2} - 5)^{2} &= 24 \\ x^{4} - 10x^{2} + 1 &= 0 \end{aligned} \]

It is minimal because \(\mathbf{Q}(\sqrt{2} + \sqrt{3}) = \mathbf{Q}(\sqrt{2}, \sqrt{3})\) has degree 4 over Q, so the minimal polynomial has degree 4.

That rules out the cubic and the quadratic. \(x^{4} - 6x^{2} + 1\) has roots \(\pm\sqrt{2} \pm 1\), the conjugates of \(1 + \sqrt{2}\), not of \(\sqrt{2} + \sqrt{3}\).

Answer: D

The Galois group is the permutation group \(S_{3}\).

The polynomial is irreducible over Q by Eisenstein's criterion at 3: 3 divides \(-3\), 15 and \(-3\), 3 does not divide the leading coefficient 1, and 9 does not divide the constant \(-3\). So the group is a transitive subgroup of \(S_{3}\), either \(A_{3} \cong \mathbf{Z}_{3}\) or \(S_{3}\).

The derivative \(3x^{2} - 6x + 15 = 3[(x - 1)^{2} + 4]\) is always positive, so f is increasing and has exactly one real root and two complex roots. Complex conjugation then swaps two roots and fixes one, a transposition, which is not in \(A_{3}\). Hence the group is \(S_{3}\). (Equivalently, the discriminant is \(-9612 \lt 0\), not a square.)

\(\mathbf{Z}_{4}\) and \(A_{4}\) cannot be Galois groups of a cubic, since the group of a cubic sits inside \(S_{3}\), of order 6.

Answer: A

There is no subfield with 16 elements.

The field with 64 elements is \(\mathbf{F}_{2^{6}}\). Its subfields are exactly the fields \(\mathbf{F}_{2^{d}}\) with d dividing 6, that is \(d = 1, 2, 3, 6\), giving subfields of 2, 4, 8 and 64 elements.

16 = \(2^{4}\), and 4 does not divide 6, so \(\mathbf{F}_{16}\) is not a subfield. (Its multiplicative group would have order 15, which does not divide 63.)

The fields with 2, 4 and 8 elements all occur, so those options are wrong.

Answer: D

The statement that need not be true is that \(\mathbf{F}_{q}\) is algebraically closed; in fact it never is.

For any finite field, the polynomial \(\prod_{a \in \mathbf{F}_{q}} (x - a) + 1\) takes the value 1 at every element, so it has no root in \(\mathbf{F}_{q}\). A finite field therefore cannot be algebraically closed.

The other three are standard facts. The order of a finite field is a prime power \(p^{n}\). There is an irreducible polynomial of every degree n over \(\mathbf{F}_{q}\), because the extension \(\mathbf{F}_{q^{n}}\) exists and is simple. And the multiplicative group of any finite field is cyclic.