Base field or ring

The answer changes with the field or ring you are working over.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option D

  The rational canonical form requires the characteristic polynomial to split

  Why this is wrong, and why it looks right

  This confuses rational canonical form with Jordan form. RCF is built from invariant factors via companion matrices and exists over any field with no factorization required; splitting the characteristic polynomial is only needed to reach Jordan form.

  From this question

  Let A, B be n×n matrices over $\mathbb{Q}$. Which of the following are true?

  Part C · free

• Option B

  2

  Why this is wrong, and why it looks right

  False. ±1 are the obvious two, but they are not the only ones once the modulus is composite. SymPy exhibits others directly: $29^{2} \equiv 1$ and $34^{2} \equiv 1 (mod 105)$. Expecting exactly two square roots of 1 is the habit carried over from fields, where $x^{2} - 1 = (x-1)(x+1)$ has at most two roots — an argument that needs the absence of zero divisors.

  From this question

  The number of solutions of the equation $x^{2} = 1$ in the ring $\mathbb{Z}/105\mathbb{Z}$ is

  See the full question

  Part B · june 2023 · free

• Option B

  $A^{2} = -I$

  Why this is wrong, and why it looks right

  False over $\mathbb{R}: A^{2}=-I$ forces the minimal polynomial to divide $x^{2}+1$, which has no real roots, so A has no real eigenvectors at all. E.g. the 90° rotation matrix satisfies $A^{2}=-I$ but is only diagonalizable over $\mathbb{C} ($eigenvalues ±i).

  From this question

  Let A be a real n×n matrix. Which of the following imply that A is diagonalisable over $\mathbb{R}$?

  Part C · free

• Option B

  F(x) has exactly four real roots.

  Why this is wrong, and why it looks right

  False. F is a quartic, so it has four roots counted with multiplicity over $\mathbb{C}$ — but the question asks about real roots. At C = +1 the roots are all complex (SymPy returns $\pm\sqrt{2}/2 \pm i\sqrt{6}/2$, none real). Degree bounds the number of real roots; it never guarantees them.

  From this question

  Let f(x) be a cubic polynomial with real coefficients. Suppose that f(x) has exactly one real root and that this root is simple. Which one of the following statements holds for ALL antiderivatives F(x) of f(x)?

  See the full question

  Part B · december 2023 · free

• Option A

  A does not have any non-zero invariant subspace in $\mathbb{C}^{n}$.

  Why this is wrong, and why it looks right

  False over $\mathbb{C}$, and this is where the field matters. $\mathbb{C}$ is algebraically closed, so the characteristic polynomial has a root, A has an eigenvector, and its span is a 1-dimensional invariant subspace. Over $\mathbb{R}$ the statement would be much closer to true — SymPy gives the rotation [[0,−1],[1,0]] eigenvalues ±i, so it has no real eigenvector and no 1-dimensional real invariant subspace. Read the field before reaching for a counterexample.

  From this question

  Let A be an n × n matrix with complex entries. If n ≥ 4, which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option C

  All eigenvalues of A are real.

  Why this is wrong, and why it looks right

  False. Nothing forces the eigenvalues of a complex matrix to be real — SymPy gives iI$_{2}$ the single eigenvalue i with multiplicity 2. Real eigenvalues are guaranteed for real symmetric matrices, and that hypothesis is absent here.

  From this question

  Let A be an n × n matrix with complex entries. If n ≥ 4, which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option B

  The Galois group of K over $\mathbb{Q}(\omega)$ is the symmetric group $S_{3}$.

  Why this is wrong, and why it looks right

  False on order grounds. Once $\omega$ is in the base field, $[K : \mathbb{Q}(\omega)] = 6/2 = 3$, so the Galois group has order 3 — it cannot be $S_{3}$, which has order 6. Enlarging the base field can only shrink the Galois group.

  From this question

  Let $f(X) = X^{3} - 2 \in \mathbb{Q}[X]$ and let $K \subset \mathbb{C}$ be the splitting field of f(X) over $\mathbb{Q}$. Let $\omega =$ $e^{2\pi{}i/3}$. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option C

  The Galois group of K over $\mathbb{Q}$ is $\mathbb{Z}/3\mathbb{Z}$.

  Why this is wrong, and why it looks right

  False. |Gal$(K/\mathbb{Q})$| $= [K : \mathbb{Q}] = 6$, so it cannot be $\mathbb{Z}/3\mathbb{Z}$. It is also non-abelian: complex conjugation and the 3-cycle on the roots do not commute. This option and (B) are the same confusion in mirror image — each attaches the right group to the wrong base field.

  From this question

  Let $f(X) = X^{3} - 2 \in \mathbb{Q}[X]$ and let $K \subset \mathbb{C}$ be the splitting field of f(X) over $\mathbb{Q}$. Let $\omega =$ $e^{2\pi{}i/3}$. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option A

  A is diagonalisable over $\mathbb{R}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  A is not diagonalisable over $\mathbb{C}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  There exists a 5 × 5 real matrix whose minimal polynomial is $(x^{2} + x + 1)^{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  V has a unique basis.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  V has precisely two bases.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  not diagonalisable over $\mathbb{C}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  Q(x, y, z) = 0 for some (x, y, z) ≠ 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  S has $\varphi(2025)$ elements, where $\varphi$ denotes the Euler $\varphi-$function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  S has $\varphi(2025)$ generators, where $\varphi$ denotes the Euler $\varphi-$function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  Both A and B are diagonalizable over $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  Neither A nor B is diagonalizable over $\mathbb{C}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  For all prime numbers p, f(X) mod p is irreducible in $(\mathbb{Z}/p\mathbb{Z})[X]$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  S is an empty set.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  Let A = [[3, 3], [1, 2]]. Then there is a matrix B with integer entries such that AB $= I_{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option B

  GL$_{2}(\mathbb{R})$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  2

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  3

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  5

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Differentiable ⇒ continuously differentiable
• Same characteristic polynomial ⇒ similar
• Diagonalisable ⇒ invertible
• Real matrix with real eigenvalues is diagonalisable
• Every real symmetric matrix is positive definite if det > 0
• Irreducible over ℤ ⇒ irreducible mod every prime

Other mistakes that cost marks