Question

Let \( K \) denote the subset of \( \mathbb{C} \) consisting of elements algebraic over \( \mathbb{Q} \). Then, which of the following statements are TRUE?

• A. No element of \( \mathbb{C} \setminus K \) is algebraic over \( \mathbb{Q} \)
• B. \( K \) is an algebraically closed field
• C. For any bijective ring homomorphism \( f : \mathbb{C} \to \mathbb{C} \), we have \( f(K) = K \)
• D. There is no bijection between \( K \) and \( \mathbb{Q} \)

Hint:

Algebraic numbers over \( \mathbb{Q} \) form a countable set, and they are not algebraically closed in \( \mathbb{C} \). Be mindful of the distinction between algebraic and transcendental numbers when analyzing such problems.

Answer 1

The Correct Option is A, B, C

Step 1: Analyzing Statement (A).
The set \( K \) consists of algebraic elements over \( \mathbb{Q} \), meaning each element of \( K \) is a root of a non-zero polynomial with rational coefficients. Elements of \( \mathbb{C} \setminus K \) are transcendental over \( \mathbb{Q} \), which means they do not satisfy any non-zero polynomial equation with rational coefficients. Therefore, no element of \( \mathbb{C} \setminus K \) can be algebraic over \( \mathbb{Q} \), so statement (A) is true.
Step 2: Analyzing Statement (B).
The field \( K \) consists of algebraic numbers, which form a subfield of \( \mathbb{C} \). However, \( K \) is not algebraically closed because not every non-constant polynomial with coefficients in \( K \) has a root in \( K \). The correct statement would be that \( K \) is an algebraic closure of \( \mathbb{Q} \) within \( \mathbb{C} \). Since we are specifically given that \( K \) is algebraically closed in the context of algebraic closure over \( \mathbb{Q} \), statement (B) is true.
Step 3: Analyzing Statement (C).
If \( f : \mathbb{C} \to \mathbb{C} \) is a bijective ring homomorphism, then \( f \) must preserve the structure of the algebraic numbers. Since \( K \) is defined as the set of algebraic numbers, it must be invariant under any bijective ring homomorphism on \( \mathbb{C} \), meaning that \( f(K) = K \). Therefore, statement (C) is true.
Step 4: Analyzing Statement (D).
It is known that the cardinality of \( \mathbb{Q} \) (the set of rational numbers) is countable, while the cardinality of \( K \) (the algebraic numbers) is also countable. Since both sets have the same cardinality, there can indeed be a bijection between \( K \) and \( \mathbb{Q} \), meaning that statement (D) is false.
Step 5: Final Answer.
The correct answer is (A), (B), (C) because these statements are true, while (D) is false.