# Can someone explain tame and wild ramification in cubic integer rings?

From my slow study of quadratic integer rings, it seems that there is only one "level" of ramification in those, so no one bothers to say if the ramification is "tame" or "wild," but I have seen those terms in regards to cubic algebraic integers.

To try to understand this, I thought I'd look at $\mathbb{Z}[\root 3 \of {12}]$. Obviously $2$ and $3$ are ramified. But then I ran into the problem that I can't factor either of those numbers as readily as I can in a ring like $\mathbb{Z}[\sqrt{6}]$ or $\mathbb{Z}[\sqrt{30}]$ (I don't yet dare try to make sense of cubic integer rings not having unique factorization).

Although I do want to factor $2$ and $3$ in $\mathbb{Z}[\root 3 \of {12}]$, I can determine without factoring the "level" of their ramification, right? And if I make that determination, does it provide any clues for factoring ramified primes?

You cannot conclude so easily that $2$ and $3$ are ramified. Write $K=\mathbb{Q}(\sqrt{12})$ and $R=\mathbb{Z}[\sqrt{12}]$. Although $2$ and $3$ clearly divide the discriminant $\Delta(R)=\Delta(X^3-12)=-972=-2^4\cdot 3^5$, the relation

$$\Delta(R)=[\mathcal{O}_K:R]^2\cdot\Delta(\mathcal{O}_K)$$ together with the observation that $2|[\mathcal{O}_K:R]$ if and only if $R$ is singular over $2$, shows that $2$ doesn't divide $\Delta(\mathcal{O}_K)$ if it happens to be the case that $4|[\mathcal{O}_K:R]$. You can prove that this does not happen and that in fact $[\mathcal{O}_K:R]=2$, but this is not immediately clear. However, since $\mathrm{ord}_3(\Delta(R))$ is odd the relation implies immediately that $3|\Delta(\mathcal{O}_K)$, so that is easier.

That aside, you can use some indirect reasoning to determine how $2$ and $3$ factor in $\mathcal{O}_K$. Note that since $R$ is singular over $2$ you don't have factorisation of $2$ in $R$ and you really need to pass to its integral closure $\mathcal{O}_K$.

For example, using the fact that $[\mathcal{O}_K:R]=2$ we see that $\mathrm{ord}_3(\Delta(\mathcal{O}_K))=5$. Now since $3$ ramifies in $\mathcal{O}_K$ we have either $3\mathcal{O}_K=\mathfrak{p}^3$ for a prime $\mathfrak{p}$ of norm $3$ or $3\mathcal{O}_K=\mathfrak{p}^2\mathfrak{q}$ for primes $\mathfrak{p},\mathfrak{q}$ both of norm $3$. If the second case would hold then we have tame ramificiation of $\mathfrak{p}$, and we see that

$$\mathrm{ord}_{\mathfrak{p}}(\mathfrak{D}_K)=e(\mathfrak{p}/3)-1=1.$$ As $\mathfrak{p}$ is also the only ramified prime over $3$ and the different $\mathfrak{D}_K$ has ideal norm $|\Delta_K|$, we are forced to conclude that $\mathrm{ord}_3(\Delta_K)=1$, which is a contradiction. It follows that $3$ is totally ramified in $\mathcal{O}_K$, and we see that we have wild ramification over $3$.

This technique unfortunately doesn't work for $2$, but it does for any prime $p\neq 2,3$. You'll see that such a $p$ is totally ramified if and only if $\mathrm{ord}_p(\Delta(\mathcal{O}_K))=2$.

To deal with $2$ you use Galois theory or factor $X^3-12$ over $\mathbb{Q}_2$. I'll illustrate the first. Since $\Delta(X^3-12)$ is not a square in $\mathbb{Q}$ it follows that the splitting field $N$ of $X^3-12$, i.e. the normal closure of $K$, is of degree $6$ over $\mathbb{Q}$ with Galois group $S_3$. Since $\Delta(X^3-12)$ is a square in $N$ but not in $\mathbb{Q}$ we obtain the unique quadratic subfield of $N$, which we know must exist by the Galois correspondence. It is $\mathbb{Q}(\sqrt{-2^4\cdot 3^5})=\mathbb{Q}(\sqrt{-3})$.

Now $2$ is unramified (in fact inert) in $\mathbb{Q}(\sqrt{-3})$, so $N/\mathbb{Q}(\sqrt{-3})$ must ramify over $2$, for else $N/\mathbb{Q}$ would be unramified over $2$. Now since $N/\mathbb{Q}(\sqrt{-3})$ is a Galois extenion of prime degree, we must have total ramification over $2$ in $N/\mathbb{Q}(\sqrt{-3})$; this is a consequence of the ram-rel identity $efg=3$ for the extension $N/\mathbb{Q}(\sqrt{-3})$. But then for the ram-rel relation $e_2f_2g_2=6$ in $N/\mathbb{Q}$ we see $2|f_2$ and $3|e_2$, so that $(e_2,f_2,g_2)=(3,2,1)$. From multiplicativity of the ramification index in the tower $N/K/\mathbb{Q}$ it now follows easily that $2$ is totally ramified in $K$, hence we have tame ramification over $2$.

Hope this helps!

EDIT. Kummer-Dedekinds theorem gives an explicit description of the primes of $K$ when $\mathcal{O}_K$ is monogenic, i.e. of the form $\mathbb{Z}[x]$ for some $x\in\mathcal{O}_K$. If this is not the case then finding explicit generators for the primes can be harder (or I'm simply not aware of the techniques).

To try to find such an $x$ in this case, we find a $\mathbb{Z}$-basis for $\mathcal{O}_K$. It is not hard to see that $\{1,\alpha,\beta\}$, with $\alpha=\sqrt{12}$ and $\beta=\alpha^2/2$ is such a basis. If now $x=a+b\alpha+c\beta$ is any element of $\mathcal{O}_K$, we have a $\mathbb{Z}$-linear map $T:\mathcal{O}_K\to\mathcal{O}_K$, which sends $1\mapsto 1$, $\alpha\mapsto x$ and $\beta\mapsto x^2$. Then $\mathrm{im}(T)=\mathbb{Z}[x]$, and $T$ is an isomorphism precisely when $\det(T)=\pm 1$. We try to find $a,b,c\in\mathbb{Z}$ such that this is the case. For this we need to expand $x^2$ in terms of the basis $\{1,\alpha,\beta\}$. From the relations $\alpha^2=2\beta$, $\beta^2=3\alpha$ and $\alpha\beta=6$ we deduce $$x^2=a^2+b^2\alpha^2+c^2\beta^2+2ab\alpha+2ac\beta+2bc\alpha\beta\\ =a^2+12bc+(2ab+3c^2)\alpha+2(ac+b^2)\beta$$ Thus the matrix representing $T$ with respect to the basis $\{1,\alpha,\beta\}$ takes the form $$\begin{pmatrix} 1 & a & a^2+12bc \\ 0 & b & 2ab+3c^2 \\ 0 & c & 2(ac+b^2) \end{pmatrix},$$ which has determinant $2b(ac+b^2)-c(2ab+3c^2)=2b^3-3c^3$. This equals $-1$ for $(a,b,c)=(0,1,1)$, and hence $\mathcal{O}_K=\mathbb{Z}[\alpha+\beta]$. The minimal polynomial of $x=\alpha+\beta$ equals $X^3-18X-30$. Using Kummer-Dedekind one now quickly finds that $2\mathcal{O}_K=\mathfrak{p}_2^3$ for $\mathfrak{p}_2=(2,\alpha+\beta)$, and that $3\mathcal{O}_K=\mathfrak{p}_3^3$ for $\mathfrak{p}_3=(3,\alpha+\beta)$.

• Since $\mathbb{Z}[\sqrt{12}] =\mathbb{Z}[x]/(x^3-12)$ let $f_p(x) =x^3 - 12 \bmod p$, ie. $f_p(x) \in \mathbb{F}_p[x]$. $f_2(x) = x^3$ is clearly ramified as well as $f_3(x) = x^3$. Jun 5, 2017 at 23:48
• The prime ideals above $p$ are of the form $(p,f_{p,j}(\sqrt{12}))$ where $f_{p,j}(x)$ is an irreducible factor of $f_p(x)$ Jun 5, 2017 at 23:50
• No the division with remainder of $X^3-12$ with respect to $X$ is $12$, which is in $2^2\mathbb{Z}[X]$, hence $2$ is singular in $\mathbb{Z}[\sqrt{12}]$. Your reasoning only works when $2$ does not divide the index $[\mathcal{O}_K:\mathbb{Z}[\sqrt{12}]]$, which is not the case. Jun 5, 2017 at 23:55
• And yes those are indeed the prime ideals, but when p divides the index $[\mathcal{O}_K:\mathbb{Z}[\sqrt{12}]]$, it is not the case that $p\mathbb{Z}[\sqrt{12}]$ factors as a product of prime ideals. Jun 5, 2017 at 23:59

It is a simple and well known fact that if $f$ is an Eisenstein polynomial for the prime number $p$, then $p$ is completely ramified in the extension generated by a root of $f$. Clearly $f(x) = x^3 - 12$ is Eisenstein for $3$, and since ${\mathbb Q}(\sqrt{12}) = {\mathbb Q}(\sqrt{18})$ and $g(x) = x^3-18$ is Eisenstein for $p = 2$, the prime $2$ is also completely ramified.

According to pari, the generators of the prime ideals are $2 + \sqrt{12} + \sqrt{18}$ for the prime ideal above $2$ and $3 + \sqrt{12} + \sqrt{18}$ for the prime ideal above $3$. This can be verified by hand in principle: cubing these elements must yield, up to a small power of the fundamental unit $\varepsilon = 1 + 3\sqrt{12} - 3\sqrt{18}$, the elements $2$ and $3$.

In fact we find $(2 + \sqrt{12} + \sqrt{18})^3 = 2(55 + 24\sqrt{12} + 21\sqrt{18}) = 2/\varepsilon$ and $(3 + \sqrt{12} + \sqrt{18})^3 = 3/\varepsilon$.