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 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 $34^2\equiv 1(mod105)$. 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

  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)?

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  Part B · december 2023

• 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?

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  Part C · december 2023

• 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

• 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?

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  Part B · december 2023

• 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

• Option A

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

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  Let A = [[3, 3], [1, 2]]. Then there is a matrix B with integer entries such that AB $=I_2$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  V has a unique basis.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  V has precisely two bases.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

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