Say we have an algebraic field with an infinite amount of units. If one multiplies two units one obtains another unit. In some cases, all units are powers of just one unit ( that's the fundamental unit, correct?) How would one go about proving the existence of a fundamental unit in such a given field ( say a cyclotomic field or real quadratic field). And if there is no such unit, maybe there are several, as in two, three, four, etc. fundamental units such that all units are powers of these units multiplied together? Again, how would one prove such a thing?

Edit: David Loeffler, thank you very much for your explanation it was very helpful. Just one detail to mention - your second unit, the one you call $u_2$, factors into $ (2^{1/2}-1)=(2^{1/4} -1)(2^{1/4}+1) $, so wouldn't it be more elegant to say that $(2^{1/4} -1)$ and $(2^{1/4}+1) $, your $u_1$, are your two ''fundamental'' units?

Edit 2: You gave an example with a fundamental unit, you said you were fond of this field... can you prove that it is indeed the fundamental unit? I am interested in the method of proof required to prove that 1) a fundamental unit exists 2) that a certain algebraic integer is indeed the fundamental unit. Again thank you.

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    $\begingroup$ Look up Dirichlet's unit theorem. $\endgroup$ Jan 14, 2015 at 16:44
  • $\begingroup$ I have... it's just I'm not familiar with all the technical language involved I hoped someone could provide an explanation I can understand easier $\endgroup$ Jan 14, 2015 at 17:05
  • $\begingroup$ What in particular confuses you, and what kind of background do you have? It is easier to address particular issues than try to explain the full proof. $\endgroup$ Jan 14, 2015 at 17:48
  • $\begingroup$ Could you give an example of a field with exactly one non-trivial fundamental unit and prove that it is indeed the fundamental unit? $\endgroup$ Jan 14, 2015 at 21:54

1 Answer 1


Let me answer your question as best I can; but I'll start by correcting a misconception in your question.

It actually never happens that "all units are powers of just one unit" except when the unit group is finite. The reason this happens is because $-1$ gets in the way! Suppose there's a unit $u$ such that every other unit $v$ can be written as $u^n$ for some $n \in \mathbf{Z}$. But $v = -1$ is a perfectly good unit, so we must be able to write $u^n = -1$; so this equation forces $u$ to be a root of unity, and in particular it can't generate an infinite group.

When the unit group is infinite, the best we can hope for is that there is a unit $u$ such that every unit $v$ is $u^n z$ where $z$ is a root of unity. This is precisely what happens for real quadratic fields: the fundamental unit $u$ has the property that every unit is $\pm u^n$ for some $n \in \mathbf{Z}$ (and you can't get rid of the $\pm 1$ there).

When $K$ has bigger degree you need more than one unit, and Dirichlet's unit theorem gives a precise formula for how many units you need. If the number field $K$ is generated by a single algebraic number $\theta$, then you look at the minimal polynomial of $\theta$ and count how many real and non-real roots it has: it will have $r$ real roots and $s$ conjugate pairs of non-real roots, for some $r$ and $s$. Dirichlet's theorem says that the smallest collection of units you need to get within a root of unity of every unit in $K$ is of size $r + s - 1$.

For instance, I'm particularly fond of the field $K = \mathbf{Q}(\sqrt[3]{2})$. The minimal polynomial of $\sqrt[3]{2}$ is $X^3 - 2$ which has one real root and two conjugate non-real roots; so in this case one unit suffices and the field $K$ does have a fundamental unit, which turns out to be $\sqrt[3]{2} - 1$.

On the other hand, the field $L = \mathbf{Q}(\sqrt[4]{2})$ corresponds to $X^4 - 2$, which has two real roots and one pair of conjugate roots; so we'll need $2 + 1 - 1 = 2$ units to generate everything -- there's no "fundamental unit" for $L$. There are computer programs for calculating in number fields, and my computer took approximately 0.07 seconds to tell me that if we take $u_1 = \sqrt[4]{2} + 1$ and $u_2 = \sqrt{2} - 1$ then every unit of $L$ is of the form $\pm u_1^a u_2^b$ for some $a, b \in \mathbf{Z}$.

PS: I will try and answer your updated questions.

You ask whether it wouldn't be better to take $\{ \sqrt[4]{2} + 1, \sqrt[4]{2} - 1\}$ rather than $\{ \sqrt[4]{2} + 1, \sqrt{2} - 1\}$ as "fundamental units". This is entirely a matter of taste: there's no really 'best' set to take, and lots of sets will work equally well. When $r + s -1$ is 1, then a fundamental $u$ will be unique up to replacing $u$ with $\omega \cdot u^{\pm 1}$ for $\omega$ one of the finite set of roots of unity in the field (usually just $\pm 1$); but as soon as you step to larger degrees there'll be an infinite amount of choice and it is a rather fruitless exercise to try and single out any one choice which is 'best'.

As for actually computing units in practice: the bible for such computations is Henri Cohen's book "A Course in Computational Algebraic Number Theory".

  • $\begingroup$ Please see the edits above ( I write this because I think you aren't notified by just an edit I need to write a comment) $\endgroup$ Jan 16, 2015 at 15:34

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