Consider the number field $K=\mathbb{Q}(\sqrt[4]{24})$. Then one can show that $\mathcal{O}_{K}=\mathbb{Z}[\alpha,\beta]$ where $\alpha=\sqrt[4]{24}$ and $\beta=\frac{\alpha^{3}}{4}$. The corresponding minimal polynomials are given by
$f_{\alpha}=X^4-24$ and $f_{\beta}=X^4-54$.
Then how does one factor the prime ideals in such a non-monogenic ring of integers. For monogenic ring of integers it can be done quite easily by using the Kummer-Dedekind Theorem.
For instance how would one factorise $5\mathcal{O}_{K}$? If we view the corresponding ideal in the subrings $\mathbb{Z}[\alpha]$ and $\mathbb{Z}[\beta]$, then the prime is actually inert. So my first thoughts were that it will also be inert in $\mathcal{O}_{K}$. But after I used some sage code I found the following factorisation
$5\mathcal{O}_{K}=(5,\frac{\alpha^2}{2}+1)(5,\frac{\alpha^2}{2}-1)$.
Any help would be appreciated!
Best Answer
The Kummer-Dedekind theorem does not require $\mathcal O_K = \mathbf Z[\alpha]$. If $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer with a minimal polynomial $f(x) \in \mathbf Z[x]$, then for every prime $p$ not dividing the index $[\mathcal O_K:\mathbf Z[\alpha]]$, the factorization of $p\mathcal O_K$ resembles that of $f(x) \bmod p$. Even if you're not sure about the value of the actual index $[\mathcal O_K:\mathbf Z[\alpha]]$, that number is a factor of the very computable disciminant ${\rm disc}(f) = {\rm disc}(\mathbf Z[\alpha])$, so you can apply the theorem to all primes not dividing ${\rm disc}(f)$ while remaining unsure if it would be valid to apply it to some of the prime factors of ${\rm disc}(f)$ (settling that would require further study of the particular situation).
The reason the theorem is called "Kummer-Dedekind" is because of the role each one played, Kummer in the 1830s and Dedekind many decades later:
Kummer only treated $\mathcal O_K = \mathbf Z[\alpha]$, and that itself is being generous since he did not have a general concept of algebraic integer. He only worked with rings of integers in cyclotomic fields, where the full ring of integers very fortunately has the form $\mathbf Z[\alpha]$ and thus Kummer could succeed without realizing the true difficulty of the general case (non-monogenic rings of integers).
Dedekind came up with the general notion of algebraic integers and he proved that Kummer's theorem still works at all primes $p$ such that $p \nmid [\mathcal O_K:\mathbf Z[\alpha]]$. At first he hoped that even if $p \mid [\mathcal O_K:\mathbf Z[\alpha]]$ he could find a $\beta$ (depending on $p$) such that $p \nmid [\mathcal O_K:\mathbf Z[\beta]]$, thus allowing him to factor each $p$ by turning that into a problem of factoring a polynomial mod $p$. That hope was dashed when he discovered the first example of a non-monogenic ring of integers ($K = \mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$, for which $[\mathcal O_K:\mathbf Z[\alpha]] = 2$ and in fact $2 \mid [\mathcal O_K:\mathbf Z[\beta]]$ for all $\beta \in \mathcal O_K - \mathbf Z$).
Example. $K = \mathbf Q(\sqrt[4]{24})$. The discriminant of $x^4 - 24$ is $-2^{17}3^3$, so for each prime $p$ other than $2$ and $3$, the way $(p)$ factors is like the way $x^4 - 24 \bmod p$ factors over $\mathbf F_p$. What about $p = 2$ and $p = 3$? The polynomial $x^4 - 24$ is Eisenstein at $3$, so $3 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ and therefore the Kummer-Dedekind theorem says the factorization of $(3)$ resembles the factorization of $x^4 - 24 \equiv x^4 \bmod 3$, so $3$ is totally ramified in $\mathcal O_K$. To treat $p = 2$, we turn instead to $x^4 - 54$, which also has a root $\sqrt[4]{54}$ that is in $\mathcal O_K$ and generates $K$. Since $x^4 - 54$ is Eisenstein at $2$, we have $2 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$ and thus the Kummer-Dedekind theorem tells us that $(2)$ factors in the same way as $x^4 - 54 \equiv x^4 \bmod 54$. So $2$ is totally ramified in $\mathcal O_K$.
I am not sure why you are saying "$5$ is inert" in $\mathbf Z[\sqrt[4]{24}]$ or $\mathbf Z[\sqrt[4]{54}]$. You must have made a typographical error when coding in Sage or you misinterpreted the Sage output. Modulo $5$, $$ x^4 - 24 = x^4 - 4 = (x^2 + 2)(x^2-2) $$ and $$ x^4 - 54 = x^4 - 4 = (x^2+2)(x^2-2). $$ These are consistent with each other (both having two irreducible factors of degree $2$), as they must be since $5$ does not divide either index $[\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ or $[\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$. Here we have an "accident" that they factor mod $5$ literally in the same way because $x^4 - 24 \equiv x^4 - 54 \bmod 5$. In any case, the Kummer-Dedekind theorem can be used with either $x^4 - 24$ or $x^4 - 54$ to factor $(5)$ and they both tell us that $(5) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ are prime ideals with norm $25$: $$ \mathfrak p = (5,\sqrt[4]{24}+2) = (5,\sqrt[4]{54}-2), \ \ \ \mathfrak q = (5,\sqrt[4]{24}-2) = (5,\sqrt[4]{54}+2), $$ where I am using positive fourth roots of $24$ and $54$ with the standard real embedding so that $\sqrt[4]{24}\sqrt[4]{54}$ is $6$ on the nose (as opposed to being $-6$). When we declare $\mathfrak p$ to be the prime ideal $(5,\sqrt[4]{24}+2)$, modulo $\mathfrak p$ we have $\sqrt[4]{24}\sqrt[4]{54} = 6 \equiv 1 \bmod \mathfrak p$, so from $\sqrt[4]{24} \equiv -2 \equiv 3 \bmod \mathfrak p$ we must have $\sqrt[4]{54} \equiv 2 \bmod \mathfrak p$ as opposed to $\sqrt[4]{54} \equiv 3 \bmod \mathfrak p$.
I don't really need real numbers or positivity, but you have to be careful with $\sqrt[4]{24}$ and $\sqrt[4]{54}$ in an abstract field $\mathbf Q(\gamma)$ where $\gamma^4 = 24$ because this field has two fourth roots of $24$ and two fourth roots of $54$. If you use fourth roots of both numbers together then you need to relate them to each other in a definite way to avoid making miscalculations. Since $6/\gamma$ is a fourth root of $54$, and its product with $\gamma$ is $6 \equiv 1 \bmod \mathfrak p$, we can say $$ \mathfrak p = (5,\gamma+2) = (5,6/\gamma-2), \ \ \ \mathfrak q = (5,\gamma-2) = (5,6/\gamma+2). $$ It would be incorrect to say $\mathfrak p = (5,\gamma+2) = (5,6/\gamma+2)$ because then modulo $\mathfrak p$ we'd have $\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$ as well as $6/\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$, so $\gamma \equiv 2 \bmod \mathfrak p$, and such congruences for $\gamma \bmod \mathfrak p$ (congruent to both $3$ and $2$) would be inconsistent.