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As you already know, given a squarefree integer $d > 1$, the ring of algebraic integers of $\textbf Q(\sqrt d)$ contains irrational algebraic integers of degree $2$ but is purely real. Given negative squarefree $d$, the ring of algebraic integers of $\textbf Q(\sqrt d)$ contains complex numbers. These are generally called "real quadratic rings" and "imaginary quadratic rings."

Now, with a squarefree integer $d > 1$, the ring of algebraic integers of $\textbf Q(\root 4 \of d)$ also contains irrational algebraic integers of degree $4$ as well as of degree $2$, and of course rational integers, but is purely real. Likewise for negative squarefree $d$, the ring of algebraic integers of $\textbf Q(\root 4 \of d)$ contains complex numbers.

I am aware that there are other kinds of numbers that, adjoined to $\textbf Q$, generate rings of algebraic integers of degree $4$, like $\zeta_8$ and $\zeta_{12}$, to name just two. I'm not concerned about those for now.

There is the related question What kinds of algebraic integers are of degree $4$?, but the answerers (including myself) were so intent on nitpicking the ancillary details of the asker's question, that there was no terminology explained.

Terminology is the essence of my question. If $d > 1$ is a squarefree integer, what do you call $\textbf Q(\root 4 \of d)$? And if $d$ is a negative and square integer, what do you call $\textbf Q(\root 4 \of d)$?

Bill Thomas
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    "$\mathbf{Q}(\sqrt[4]{d})$ [for $d > 0$] ... is purely real" is misleading. The corresponding field has signature $(2,1)$, that is, of the four complex embeddings of this field, only two are real. (The polynomial $x^4 - d$ has two real roots and two purely imaginary roots for $d > 0$.) I think the term "pure quartic number field" would be widely recognized as a name for these fields (with no restriction on the sign of $d$). – The Piper Jun 27 '19 at 19:44
  • You wouldn't impose the condition that $d$ is squarefree either; why rule out fields like $\mathbf{Q}(\sqrt[4]{24}) = \mathbf{Q}(\sqrt[4]{54})$? (fourth power free, perhaps?) – The Piper Jun 27 '19 at 19:47
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    @ThePiper Maybe this is obvious to you and Bill, but I'm confused. Let's say $d$ is such that $\root 4 \of d$ is positive, real but irrational. We adjoin that number to the real rationals and we get a ring that also includes the still real though negative number $-\root 4 \of d$. But how can this also include $\pm i \root 4 \of d$? – David R. Jun 28 '19 at 21:24
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    It doesn't, it's just that the abstract field $K = \mathbf{Q}[x]/(x^4 - d)$ has $4$ embeddings into the complex numbers, two of them land inside the reals and the other two do not. So your "subfield of the reals" given by $\mathbf{Q}(\sqrt[4]{d})$ is isomorphic to $\mathbf{Q}i \sqrt[4]{d})$, so "purely real" is not great terminology. – The Piper Jun 28 '19 at 21:40

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"If $d > 1$ is a squarefree integer, what do you call $\Bbb Q(\root 4 \of d)$?" I think it is just called a quartic number field because of $$ [\Bbb Q(\sqrt[4]{d}):\Bbb Q]=4. $$ Sometimes it is also called a pure quartic number field. Of course, one could have different names for $d<0$ and $d>0$. Another type of number fields of degree $4$ are the fields $\Bbb Q(\sqrt{n},\sqrt{m})$ for suitable $m,n$, called biquadratic number field, or cyclotomic number fields $\Bbb Q(\zeta_n)$ with $\phi(n)=4$.

Dietrich Burde
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