A number is an "algebraic integer" if it is the root to a monic polynomial with integer coefficients. Artin says (Algebra, p. 411):

The concept of algebraic integer was one of the most important discoveries of number theory. It is not easy to explain quickly why it is the right definition to use, but roughly speaking, we can think of the leading coefficient of the primitive irreducible polynomials $f(x)$ as a "denominator." If $\alpha$ is the root of an integer polynomial $f(x)=dx^n+a_{n-1}x^{n-1}...$ then $d\alpha$ is an algebraic integer, because it is a root of the monic integer polynomial $x^n + a_{n-1}x^{n-1} + ... + d^{n-1}a_0$.

Thus we can "clear the denominator" in any algebraic number by multiplying it with a suitable integer to get an algebraic integer.

When I first learned of algebraic integers, I looked online and saw some hints that maybe they were used to prove the Abel-Ruffini theorem. So I put off questioning their usage for a while; I now think I understand one proof of this theorem (the one at the end of Artin's Algebra) and it has nothing to do with algebraic integers (that I can tell).

So basically: why is it important if a number is an algebraic integer? I think I understand what he's saying about the relationship between roots of integer polynomials and algebraic integers, but I fail to see why this is "one of the most important discoveries of number theory."

  • $\begingroup$ You can use algebraic integers to construct extensions of the ring of integers. The technique is far more important than the fact that "a number is an algebraic integer". $\endgroup$
    – Sasha
    Commented Oct 5, 2011 at 15:46
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    $\begingroup$ @Xodarap: Dear Xodarap, Why do you care about integers? Maybe you don't, but if you do, one reason might be because they are a very interesting subring of $\mathbb Q$, with subtle divisibility properties (and prime numbers and so on) which are lost after passing to $\mathbb Q$ (where very non-zero element is invertible). If $K$ is a finite extension of $\mathbb Q$, then its subring of algebraic integers stands in precisely the same relationship to $K$ as $\mathbb Z$ does to $\mathbb Q$. Regards, $\endgroup$
    – Matt E
    Commented Oct 5, 2011 at 16:17
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    $\begingroup$ There are instances where $\mathbb{Z}[\alpha]$ is not a unique factorization domain, for example, but where the ring of algebraic integers in $\mathbb{Q}[\alpha]$ is a unique factorization domain. For example, $\mathbb{Z}[\sqrt{-3}]$ I believe, is not a unique factorization domain, but the ring of algbraic integers in $\mathbb{Q}[\sqrt{-3}]$ is. $\endgroup$ Commented Oct 5, 2011 at 16:47
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    $\begingroup$ @Xodarap: No, but you can recover a sort of unique factorization using ideals; see Arturo's answer. $\endgroup$ Commented Oct 5, 2011 at 18:52
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    $\begingroup$ @Xodarap, not even close. For example, $\mathbb{Z}[\sqrt{3}]$ has your property, but it isn't a UFD. As Chris notes, if you take the algebraic integers in $Q\otimes R$, you don't always get a UFD, but you do get other nice properties. $\endgroup$ Commented Oct 5, 2011 at 20:19

2 Answers 2


Suppose we desire to consider as "integers" some subring $\,\Bbb I\,$ of the field of all algebraic numbers. To be a purely algebraic notion, it cannot distinguish between conjugate roots, so if $\rm\,\alpha,\alpha'$ are roots of the same polynomial irreducible over $\rm\,\Bbb Q\,,\,$ then $\rm\,\alpha\in\Bbb I\iff \alpha'\in\Bbb I\,.\,$ Furthermore we desire $\rm\,\Bbb I\cap \Bbb Q = \Bbb Z\ $ so that our notion of algebraic integer is a faithful extension of the notion of a rational integer. Now suppose that $\rm\,f(x)\,$ is the monic minimal polynomial over $\rm\,\Bbb Q\,$ of an algebraic "integer" $\rm\,\alpha\in \Bbb I\,.\,$ Then $\rm\,f(x) = (x-\alpha)\,(x-\alpha')\,(x-\alpha'')\,\cdots\,$ has coefficients in $\rm\,\Bbb I\cap \Bbb Q = \Bbb Z.\,$ Therefore the monic minimal polynomial of elements $\in\Bbb I\,$ must have coefficients $\in\Bbb Z\,.\,$ Conversely, one easily shows that the set of all such algebraic numbers contains $1$ and is closed under both difference and multiplication, so it forms a ring. Moreover, as Artin's quote shows, the quotient field of $\rm\,\Bbb I\,$ is the field of all algebraic numbers. Hence a few natural hypotheses on the notion of an algebraic integer imply the standard criterion in terms of minimal polynomials.

Because this notion of integer faithfully extends the notion of rational integers, we can employ algebraic integers to deduce results about rational integers. This method often results in great simplifications because many diophantine equations become simpler - being "linearized" - when we factor them in algebraic extension fields, e.g. see proofs about Pythagorean triples using Gaussian integers, or classical proofs of FLT for small exponents employing algebraic integers.

  • $\begingroup$ Could you elaborate on why $f\in Z[x]$? $\endgroup$
    – Xodarap
    Commented Oct 5, 2011 at 19:43
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    $\begingroup$ @Xod By hypothesis, since $\rm\:\alpha\in\mathbb I\ $ so too are all the "conjugate" roots $\rm\:\alpha',\ \alpha'',\:\cdots\:$ of the minimal polynomial $\rm\:f.\:$ Since by hypothesis $\rm\:\mathbb I\:$ is a ring, it contains all negations, sums and products of such conjugates, which includes all of the coefficients of $\rm\:f,\:$ for example $\rm\:f_0 = (-1)^n \alpha\ \alpha'\:\alpha''\cdots,\:$ and $\rm\:-f_{\,n-1} = \alpha+\alpha'+\alpha''+\cdots.\:$ By definition, $\rm\:f\in\mathbb Q[x],\:$ so the coefficients of $\rm\:f\:$ lie in $\rm\:\mathbb I\cap\mathbb Q = \mathbb Z.\qquad$ $\ $ $\endgroup$ Commented Oct 5, 2011 at 20:17
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    $\begingroup$ One could remark that in general, it may well happen that, while $\alpha$ is an algebraic number, $$\mathbb{Z}[\alpha] \cap \mathbb{Q} = \mathbb{Z}$$ (thus, closed-ness under conjugation is indeed important). Another reason to define $\mathcal{O}_K$ in the way it is defined is because $\mathbb{Z}[\alpha]$ is a finitely generated $\mathbb{Z}$-module iff $\alpha$ is an algebraic integer. $\endgroup$ Commented Jun 11, 2014 at 15:26
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    $\begingroup$ @streetcar277 Yes, that alone does not suffice, there are even simple quadratic examples, for example $\,\Bbb Z[\alpha],\,$ for $\alpha = (3-\sqrt{2})/7\ $ or $\ (3-\sqrt{-7})/4.\ \ $ $\endgroup$ Commented Jun 11, 2014 at 15:33

The notion had already appeared in work of Eisenstein, though only in passing. Algebraic integers came to the fore in the work of Dedekind and Kronecker, that in turn extended the ideas of Kummer.

If you remember Euler's attempt to prove Fermat's Last Theorem for $n=3$ and Lamé's attempted general proof, they both were implicitly assuming that you could work in a ring of the form $\mathbb{Z}[\theta]$, with $\theta$ algebraic, and have a kind of "unique factorization"; Euler worked in $\mathbb{Z}[\sqrt{3}]$, Lamé was working in $\mathbb{Z}[\zeta_p]$, where $\zeta_p$ is a primitive $p$-th root of unity.

Although they don't generally have unique factorization, as Kummer had shown, Kummer was able to rescue a certain amount of unique factorization for rings of the latter type, $\mathbb{Z}[\zeta_p]$, by using ideal numbers, and showing that every element of $\mathbb{Z}[\zeta_p]$ can be expressed uniquely as a product of actual and ideal prime numbers.

But it was not clear how to extend the ideas to arbitrary extensions of $\mathbb{Q}$. If you start with $\mathbb{Q}(\alpha)$, and try to do it with $\mathbb{Z}[\alpha]$, then things can go very wrong. Turns out that the "right" notion to use is that of algebraic integer: you want to consider all numbers in $\mathbb{Q}(\alpha)$ that are algebraic integers, and that forms your ring of "integers of $\mathbb{Q}(\alpha)$", over which you can do number theory.

Then you can recover a kind of "unique factorization", namely, "unique factorization into ideals": for every ideal $I$ of the ring $\mathscr{O}(\alpha)$ of all algebraic integers that are in $\mathbb{Q}(\alpha)$, you can express $I$ uniquely as a product $$I = \mathfrak{p}_1^{\alpha_1}\cdots \mathfrak{p}_m^{\alpha_m}$$ of prime ideals $\mathfrak{p}_i$, with the primes pairwise distinct, and the exponents positive integers, with the expression unique up to order. In many considerations, this allows you to recover "enough" unique factorization to proceed (though not in all situations). Taking an element $a\in\mathscr{O}(\alpha)$, you can consider $I=(a)$, and this gives you the kind of "factorization" that Kummer obtained for cyclotomic integers, with principal prime ideals corresponding to "actual primes", and non-principal prime ideals corresponding to "ideal primes".


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