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Not that I fully understand quadratic integer rings yet, but I've been wondering about quartic integer rings.

Which leads me to the question: what kinds of algebraic integers generate quartic integer rings? I am fairly certain only square roots of squarefree integers are algebraic integers of degree $2$. So for degree $4$, these are the numbers that I am aware of:

  • Fourth roots of squarefree integers with prime factors, like $\root 4 \of 5$, $\root 4 \of 6$, etc. (so $-i$ is not among these, despite being a fourth root of $1$).
  • Sums of two square roots of coprime squarefree integers, like $\sqrt 2 + \sqrt 3$, $\sqrt 5 + \sqrt 7$.

What am I missing?

Mr. Brooks
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    Numbers of the form $\sqrt{a + \sqrt b}$ are missing, at the very least. See this wikipedia link, and especially the image to the right for the general form of a degree-$4$ algebraic number. Set $a = 1$, and you get algebraic integers. – Arthur Jan 28 '17 at 22:40
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    You are missing simply all the roots of monic irreducible polynomials of degree $4$ with integer coefficients. – Crostul Jan 28 '17 at 22:42
  • You're also forgetting a lot of cubefree numbers that are not squarefree, e.g., $\root 4 \of{12}$ has a minimal polynomial of $x^4 - 12$. But you're right to avoid fourth roots of perfect squares, since those are simply the square roots of their square roots. – Robert Soupe Jan 30 '17 at 02:02
  • Any thoughts, Mr. Brooks, about the comments and the answer? – Gerry Myerson Jan 30 '17 at 12:04
  • Arthur's comment is very enlightening, @GerryMyerson – Mr. Brooks Jan 30 '17 at 22:00

3 Answers3

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Among other things, you are missing $~2x~=~\sqrt{\alpha-\beta}~-~\sqrt{\beta-\alpha+\dfrac2{\sqrt{\alpha-\beta}}}~,~$ which, for $~\alpha=\sqrt[3]{\dfrac12+\dfrac{\sqrt{849}}{18}}~$ and $~\beta=4~\sqrt[3]{\dfrac2{3(9+\sqrt{849})}}~,~$ is one of the solutions of $x^4-x-1=0$.

The point being, explicit formulas for algebraic integers of degree four can be very, very complicated, much more complicated than for degree two.

$($To get another version of this number, type $x^4-x-1=0$ into Wolfram Alpha, and after it gives you a numerical solution, ask it for the "exact form".$)$

Gerry Myerson
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  • Gerry, you seem to have forgotten to add dollar signs to your post... – DonAntonio Jan 28 '17 at 23:39
  • @Don, it would take a lot more than dollar signs to make that mess typeset properly. Every sqrt(849) would have to be edited into \sqrt{849}, every ^(1/3) into ^{1/3}, every 1/2 into \frac{1}{2} or {1\over2}, and so on, and so on. You're welcome to do that, if you want to, but I think my 2nd & 3rd paragraphs justify not expending the effort. – Gerry Myerson Jan 29 '17 at 04:50
  • Sounds like simple mass replace. $$x = \frac{1}{2} \sqrt((\frac{1}{2} (9 + \sqrt{849}))^{1/3}/3^{2/3} - 4 (2/(3 (9 + \sqrt{849})))^{1/3}) - \frac{1}{2} \sqrt(4 (2/(3 (9 + \sqrt{849})))^{1/3} - (\frac{1}{2} (9 + \sqrt{849}))^{1/3}/3^{2/3} + 2/ \sqrt((\frac{1}{2} (9 + \sqrt{849}))^{1/3}/3^{2/3} - 4 (2/(3 (9 + \sqrt{849})))^{1/3}))$$ – Robert Soupe Jan 29 '17 at 17:45
  • Though of course it still can be improved upon. – Robert Soupe Jan 29 '17 at 17:47
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    What is the intuitive argument for explaining those $\sqrt{\ldots +(a+\sqrt{b})^{1/3}}$ ? That $\phi(12) = 4$ ? @RobertSoupe – reuns Jan 30 '17 at 23:10
  • @user1952009 I was only addressing the issue of typesetting. It's time-consuming and annoying, but not impossible. – Robert Soupe Jan 31 '17 at 02:28
  • @Robert, I don't know what you mean by intuitive, but the way to solve a general quartic involves producing and solving a "cubic resolvent," and the cube roots come up in solving that cubic. Have a look at any treatment of the solution of the quartic. Or, you could appeal to Galois Theory: the Galois group of this quartic is $S_4$, the symmetric group on four letters; this group has a normal subgroup $A_4$ with quotient $S_4/A_4$ cyclic of order two, and $A_4$ has a normal subgroup $V$ with quotient $A_4/V$ cyclic of order three, and $V$ is abelian of order 4; [continued] – Gerry Myerson Jan 31 '17 at 02:43
  • [continued from previous comment] the cube roots come from that quotient of order three. – Gerry Myerson Jan 31 '17 at 02:44
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    Gerry, I think you meant to ping @user1952009. Still, I appreciate the explanation. – Robert Soupe Jan 31 '17 at 04:59
  • My thanks to user Lucian for editing in the formatting. – Gerry Myerson Feb 02 '17 at 22:03
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Try a nontrivial fifth root of unity, being a root of $(x^5-1)/(x-1)$. It is ultimately hopeless and not a wise use of time to try to describe all algebraic integers of degree 4.

Your description of all algebraic integers of degree 2 is incomplete also. Try $(1+\sqrt{5})/2$. There are many more examples where that came from.

KCd
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    The old man forgot he once identified $$\frac{3}{2} + \frac{\sqrt{-31}}{2}$$ as part of an alternate factorization of 10 in that ring. I remember because I was one of the people to answer that question. Guess it happens when you get to be his age. – Robert Soupe Jan 31 '17 at 05:05
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I choose not to be daunted by this question, even though in some regards it is daunting, however much that has been overplayed so far. I think that a lot of these algebraic integers of degree $4$ can be boiled down to $a + b \theta + c \theta^2 + d \theta^3$, where $a, b, c, d$ are all integers, or perhaps all rational numbers satisfying a certain condition, and $\theta$ is an algebraic integer of degree $4$.

Something tells me that it is this $\theta$ that you're actually interested in, namely, your apparent ignorance of quadratic integers like $\omega$ and $\phi$ despite your earlier demonstrated acquaintance with them.

It is for these $\theta$ that things get hairy. What I have pieced together so far, mainly from your question and from comments:

  • Fourth roots of integers, provided they are not perfect powers and not divisible by any fourth powers.
  • Sums of two square roots of coprime squarefree integers.
  • Square root of an integer plus a square root.
  • Maybe the an integer plus a square root, divided by a square root or the cube of a square root?

At least we don't have to worry about Abel's impossibility theorem at this degree.

Bill Thomas
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