Is 1/2 an algebraic integer? Which answer correctly identifies the status of 1/2 and explains why?
AYes — it satisfies 2x − 1 = 0, which has integer coefficients, so it is an algebraic integer
BNo — it is not an integer, and only integers can be algebraic integers
CNo — no monic polynomial with integer coefficients has 1/2 as a root, so the defining criterion is not met
DYes — all rational numbers are algebraic integers because they can be expressed with integer numerators and denominators
The definition requires a *monic* polynomial with integer coefficients — leading coefficient must be 1. The number 1/2 satisfies 2x − 1 = 0, but this polynomial is not monic. If you try to write a monic polynomial with 1/2 as a root, you get x − 1/2 = 0, whose coefficients are not all integers. Option A is the classic misconception: any polynomial with integer coefficients is not sufficient; the monic requirement is the whole point. The rational algebraic integers are exactly ℤ itself — rational numbers with no integer polynomial whose monic version has integer coefficients.
Question 2 Multiple Choice
In ℚ(√−3), the ring of integers 𝒪_K turns out to be ℤ[ω] where ω = (−1 + √−3)/2, not just ℤ[√−3]. This is initially surprising because ω looks like a 'half-integer.' Why does ω qualify as an algebraic integer?
ABecause ω has absolute value less than 1, placing it within the unit disk where all algebraic integers live
BBecause ω satisfies x² + x + 1 = 0, which is monic with integer coefficients, so the definition is satisfied
CBecause ω is a primitive cube root of unity and all roots of unity are automatically algebraic integers by convention
DBecause ℤ[ω] contains ℤ as a subring, and any extension of ℤ consists of algebraic integers
The answer is purely definitional: ω satisfies x² + x + 1 = 0, which is monic (leading coefficient 1) with integer coefficients (1, 1, 1). That is the complete criterion for being an algebraic integer, and ω meets it. Option C contains a true statement — roots of unity are algebraic integers — but the *reason* is that they satisfy monic integer polynomials like xⁿ − 1 = 0, not 'by convention.' Option D is false: extensions of ℤ can contain non-integers (like ℚ). The 'half-integer' appearance of ω is misleading intuition that the monic polynomial criterion overrides.
Question 3 True / False
The sum and product of two algebraic integers are always algebraic integers — that is, the set of algebraic integers in any number field forms a ring.
TTrue
FFalse
Answer: True
This is one of the key structural facts about algebraic integers. If α satisfies a monic integer polynomial of degree m and β satisfies one of degree n, then both α + β and αβ satisfy monic integer polynomials of degree at most mn. The proof uses the fact that the minimal polynomial of α + β divides the characteristic polynomial of a certain matrix constructed over ℤ. This closure under addition and multiplication is what makes 𝒪_K a ring — and it's not at all obvious from the definition alone.
Question 4 True / False
A number is an algebraic integer if and only if it is a root of some polynomial with integer coefficients.
TTrue
FFalse
Answer: False
The 'monic' requirement is essential and this statement omits it. Every algebraic integer is indeed a root of an integer polynomial, but the converse fails: 1/2 is a root of 2x − 1 = 0 (an integer polynomial) but is not an algebraic integer because no *monic* integer polynomial has 1/2 as a root. The distinction between 'integer polynomial' and 'monic integer polynomial' is precisely what separates algebraic numbers (roots of any integer polynomial, monic or not) from algebraic integers (roots of monic integer polynomials). Rational algebraic numbers that are not integers, like 1/2 or 2/3, are always excluded by the monic requirement.
Question 5 Short Answer
Explain why the monic requirement in the definition of algebraic integer is essential. What goes wrong — concretely — if we drop it and allow any polynomial with integer coefficients?
Think about your answer, then reveal below.
Model answer: If we drop the monic requirement and define 'algebraic integer' as any root of any polynomial with integer coefficients, then every algebraic number would qualify — including 1/2 (root of 2x − 1 = 0) and 1/3 (root of 3x − 1 = 0). The resulting set would just be the algebraic numbers ℚ-bar, which is a field, not a useful generalization of ℤ. The monic requirement is what ensures that the rational algebraic integers are exactly ℤ itself — giving the right generalization: just as ℤ ⊂ ℚ, so 𝒪_K ⊂ K, and the 'integers' of K are the ones satisfying the monic condition.
The monic requirement is the defining feature that makes algebraic integers a genuine generalization of ordinary integers rather than a trivial extension. It ensures that the notion of 'integer' in any number field reduces to ℤ when restricted to ℚ, and it produces a ring (closed under addition and multiplication) rather than an arbitrary subset. Without it, the theory loses its structural content and the connection to factorization and ideal theory in number fields collapses.