Algebraic integer

Examples

• The only algebraic integers that are found in the set of rational numbers are the integers. In other words, the intersection of ${\displaystyle \mathbb{Q}}$ and A is exactly ${\displaystyle \mathbb{Z}}$. The rational number ⁠a/b⁠ is not an algebraic integer unless b divides a. The leading coefficient of the polynomial bx − a is the integer b.
• The square root ${\displaystyle \sqrt{n}}$ of a nonnegative integer n is an algebraic integer, but is irrational unless n is a perfect square.
• If d is a square-free integer then the extension ${\displaystyle K=\mathbb{Q}(\sqrt{d})}$ is a quadratic field of rational numbers. The ring of algebraic integers O_K contains ${\displaystyle \sqrt{d}}$ since this is a root of the monic polynomial x² − d. Moreover, if d ≡ 1 mod 4, then the element $\frac{1}{2}(1+\sqrt{d})$ is also an algebraic integer. It satisfies the polynomial x² − x + ⁠1/4⁠(1 − d) where the constant term ⁠1/4⁠(1 − d) is an integer. The full ring of integers is generated by ${\displaystyle \sqrt{d}}$ or $\frac{1}{2}(1+\sqrt{d})$ respectively. See Quadratic integer for more.
• The ring of integers of the field ${\displaystyle F=\mathbb{Q}[\alpha ]}$, α = ³√m (where m is cube-free), has the following integral basis, writing m = hk² for two square-free coprime integers h and k: $${\displaystyle \{\begin{array}{ll}1,\alpha ,{\displaystyle \frac{{\alpha }^2\pm k^2\alpha +k^2}{3k}}& m\equiv \pm 1mod9\\ 1,\alpha ,{\displaystyle \frac{{\alpha }^2}{k}}& \text{otherwise}\end{array}}$$
• If ζ_n is a primitive nth root of unity, then the ring of integers of the cyclotomic field ${\displaystyle \mathbb{Q}({\zeta }_n)}$ is precisely ${\displaystyle \mathbb{Z}[{\zeta }_n]}$.
• If α is an algebraic integer then β = ⁿ√α is another algebraic integer. A polynomial for β is obtained by substituting xⁿ in the polynomial for α.

Finite generation of ring extension

For any α, the ring extension (in the sense that is equivalent to field extension) of the integers by α, denoted by ${\displaystyle \mathbb{Z}[\alpha ]\equiv \left\{\sum _{i=0}^n{\alpha }^i{z}_i|z_i\in \mathbb{Z},n\in \mathbb{Z}\right\}}$, is finitely generated if and only if α is an algebraic integer.

The proof is analogous to that of the corresponding fact regarding algebraic numbers, with ${\displaystyle \mathbb{Q}}$ there replaced by ${\displaystyle \mathbb{Z}}$ here, and the notion of field extension degree replaced by finite generation (using the fact that ${\displaystyle \mathbb{Z}}$ is finitely generated itself); the only required change is that only non-negative powers of α are involved in the proof.

The analogy is possible because both algebraic integers and algebraic numbers are defined as roots of monic polynomials over either ${\displaystyle \mathbb{Z}}$ or ${\displaystyle \mathbb{Q}}$, respectively.

Ring

The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is not. Thus the algebraic integers form a ring.

This can be shown analogously to the corresponding proof for algebraic numbers, using the integers ${\displaystyle \mathbb{Z}}$ instead of the rationals ${\displaystyle \mathbb{Q}}$.

One may also construct explicitly the monic polynomial involved, which is generally of higher degree than those of the original algebraic integers, by taking resultants and factoring. For example, if x² − x − 1 = 0, y³ − y − 1 = 0 and z = xy, then eliminating x and y from z − xy = 0 and the polynomials satisfied by x and y using the resultant gives z⁶ − 3z⁴ − 4z³ + z² + z − 1 = 0, which is irreducible, and is the monic equation satisfied by the product. (To see that the xy is a root of the x-resultant of z − xy and x² − x − 1, one might use the fact that the resultant is contained in the ideal generated by its two input polynomials.)

Integral closure

Every root of a monic polynomial whose coefficients are algebraic integers is itself an algebraic integer. In other words, the algebraic integers form a ring that is integrally closed in any of its extensions.

Again, the proof is analogous to the corresponding proof for algebraic numbers being algebraically closed.

Additional facts

• Any number constructible out of the integers with roots, addition, and multiplication is an algebraic integer; but not all algebraic integers are so constructible: in a naïve sense, most roots of irreducible quintics are not. This is the Abel–Ruffini theorem.
• The ring of algebraic integers is a Bézout domain, as a consequence of the principal ideal theorem.
• If the monic polynomial associated with an algebraic integer has constant term 1 or −1, then the reciprocal of that algebraic integer is also an algebraic integer, and each is a unit, an element of the group of units of the ring of algebraic integers.
• If x is an algebraic number then a_nx is an algebraic integer, where x satisfies a polynomial p(x) with integer coefficients and where a_nxⁿ is the highest-degree term of p(x). The value y = a_nx is an algebraic integer because it is a root of q(y) = a^{n − 1}
  _n p(y/a_n), where q(y) is a monic polynomial with integer coefficients.
• If x is an algebraic number then it can be written as the ratio of an algebraic integer to a non-zero algebraic integer. In fact, the denominator can always be chosen to be a positive integer. The ratio is |a_n|x / |a_n|, where x satisfies a polynomial p(x) with integer coefficients and where a_nxⁿ is the highest-degree term of p(x).
• The only rational algebraic integers are the integers. That is, if x is an algebraic integer and ${\displaystyle x\in \mathbb{Q}}$ then ${\displaystyle x\in \mathbb{Z}}$. This is a direct result of the rational root theorem for the case of a monic polynomial.