# Quotients of number fields by certain prime powers

I apologise in advance for what must be a naive question. Let $$\mathcal O_K$$ be the ring of integers of the algebraic number field $$K.$$ Let $$p$$ be a rational prime, and factorize $$(p)=\mathfrak p_1^{e_1}\cdots\mathfrak p_r^{e_r}$$ where the $$\mathfrak p_i$$ are primes in $$\mathcal O_K.$$ Let $$k_i=\mathcal O_K/\mathfrak p_i$$ be the residue fields for $$1\le i\le r.$$

I've seen a statement without proof that $$\mathcal O_K / (p)$$ is ring isomorphic to the sum of truncated polynomial rings $$\bigoplus_1^r k_i[t]/(t^{e_i}).$$ It looks like a standard result, but I can't find a proof, could anyone point out a source? I see a simple proof where $$\mathcal O_K$$ is $$p$$-monogenic (i.e. $$|\mathcal O_K:\mathbb Z[\alpha]|$$ is prime to $$p$$ for some $$\alpha\in\mathcal O_K,$$ the usual condition for Dedekind's criterion to be applied) though I would still welcome a source to check my reasoning, but I understand the result holds without that assumption.

• That is not true. Commented Mar 5, 2023 at 12:56
• Are you asking for an isomorphism as vector spaces over $\mathbb{Z}/p$, as rings, or in some other sense? Commented Mar 5, 2023 at 13:06

I remember working this out 25 years ago. The main idea is to view both rings as quotient rings of completions: $$\mathcal O_K/\mathfrak p^e \cong \widehat{\mathcal O_{\mathfrak p}}/\widehat{\mathfrak p}^e$$ and $$k[t]/(t^e) \cong k[[t]]/(t^e)$$. Then the structure of the ring of integers of local fields, in characteristic $$0$$ and characteristic $$p$$, will help us solve the problem.

For a nonzero prime ideal $$\mathfrak p$$ in $$\mathcal O_K$$ lying over $$p$$ and $$m \geq 1$$, the quotient ring $$\mathcal O_K/\mathfrak p^m$$ is unchanged up to isomorphism if we replace $$\mathcal O_K$$ by its localization at $$\mathfrak p$$ or by its completion at $$\mathfrak p$$ (and the modulus also changes to the ideal it generates in the localization or completion). Also, the ramification index $$e = e(\mathfrak p|p)$$ and residue field degree $$f = f(\mathfrak p|p)$$ are unchanged by localizing or completing at $$\mathfrak p$$.

Let $$A = \widehat{\mathcal O_{\mathfrak p}}$$ be the completion of $$\mathcal O_K$$ at $$\mathfrak p$$, so in $$A$$ we can write $$p = \pi^e u$$ for some uniformizer $$\pi$$ and $$u \in A^\times$$. Then $$\mathcal O_K/\mathfrak p^e \cong A/(\pi^e) = A/(p)$$. (In this step it's important that the exponent $$e$$ is the ramification index of $$\mathfrak p$$ over $$p$$.) The ring $$A$$ is the ring of integers of the completion $$K_\mathfrak p$$. Even though $$\mathcal O_K$$ need not be monogenic, completions are monogenic: the ring of integers of every finite extension of $$\mathbf Q_p$$ has a power basis over $$\mathbf Z_p$$, so $$A = \mathbf Z_p[\alpha]$$ for some $$\alpha \in A$$. That there is a power basis over $$\mathbf Z_p$$ is the key fact you were missing.

Let $$\alpha$$ have minimal polynomial $$F(x)$$ in $$\mathbf Z_p[x]$$, so $$F(x)$$ is irreducible over $$\mathbf Z_p$$ and $$A \cong \mathbf Z_p[x]/(F(x))$$ as rings. Viewing both sides as $$\mathbf Z_p$$-modules and computing their $$\mathbf Z_p$$-ranks shows $$[K_\mathfrak p:\mathbf Q_p] = \deg F$$, so $$\deg F = e(\mathfrak p|p)f(\mathfrak p|p) = ef.$$

Using $$F(x)$$, $$\mathcal O_K/\mathfrak p^e \cong A/(p) = \mathbf Z_p[\alpha]/(p) \cong \mathbf Z_p[x]/(p,F(x)) \cong \mathbf F_p[x]/(\overline{F}(x)).$$

The mod $$p$$ reduction $$\overline{F}(x)$$ in $$\mathbf F_p[x]$$ has a factorization into monic irreducibles. All monic irreducible factors of $$\overline{F}(x)$$ are the same, because if that were not the case then we could write $$F(x) \equiv G(x)H(x) \bmod p$$ for nonconstant monic $$G(x)$$ and $$H(x)$$ where $$\gcd(G \bmod p,H \bmod p) = 1$$ in $$\mathbf F_p[x]$$, and then by Hensel's lemma (the one about lifting relatively prime factorizations, not just about lifting a simple root) we'd get a factorization of $$F(x)$$ in $$\mathbf Z_p[x]$$ into nonconstant monic factors, which contradicts the irreducibility of $$F(x)$$ over $$\mathbf Z_p$$. So in $$\mathbf F_p[x]$$ we must have $$\overline{F}(x) = Q(x)^d$$ for some monic irreducible $$Q(x)$$ in $$\mathbf F_p[x]$$ and $$d \geq 1$$. That means $$\mathcal O_K/\mathfrak p^e \cong A/(p) \cong \mathbf F_p[x]/(Q(x)^d)$$ as rings.

We will now show $$d = e = e(\mathfrak p|p), \ \ \deg Q = f = f(\mathfrak p|p).$$ Let $$k = \mathcal O_K/\mathfrak p$$, which is the residue field of $$\mathcal O_K$$ at $$\mathfrak p$$, so $$\dim_{\mathbf F_p}(k) = f$$ by definition. Residue fields are unchanged up to isomorphism by completion, so $$k \cong A/(\pi)$$. The local ring $$A/(p) = A/(\pi^e)$$ has maximal ideal $$(\pi)/(\pi^e)$$ and residue field $$A/(\pi)$$, while the local ring $$\mathbf F_p[x]/(Q(x)^d)$$ has maximal ideal $$(Q(x))/(Q(x)^d)$$ and residue field $$\mathbf F_p[x]/(Q(x))$$. Isomorphic local rings have isomorphic residue fields, so $$k \cong \mathbf F_p[x]/(Q(x))$$. Computing $$\mathbf F_p$$-dimensions of both sides, $$f = \deg Q.$$ Returning to $$F$$, which is monic and reduces mod $$p$$ to $$Q(x)^d$$, we can now say $$\deg F = \deg \overline{F} = d\deg Q = d f$$ and we already saw $$\deg F = ef$$, so $$d = e.$$ Thus $$\mathcal O_K/\mathfrak p^e \cong A/(p) \cong \mathbf F_p[x]/(Q(x)^e), \ \ \deg Q = f.$$

The last step is to show $$\mathbf F_p[x]/(Q(x)^e) \cong k[t]/(t^e)$$ as rings, so $$\mathcal O_K/\mathfrak p^e \cong A/(p) \cong \mathbf F_p[x]/(Q^e) \cong k[t]/(t^e).$$

In fact we'll show $$\mathbf F_p[x]/(Q(x)^m) \cong k[t]/(t^m)$$ for all $$m \geq 1$$. We need $$m = e$$ only to identify these rings with $$\mathcal O_K/\mathfrak p^e$$, but the rings are isomorphic to each other for all $$m \geq 1$$. (Note $$\mathcal O_K/\mathfrak p^m$$ has characteristic $$p$$ if and only if $$\mathfrak p^m \mid p\mathcal O_K$$, forcing $$m \leq e$$, but the rings $$\mathbf F_p[x]/(Q(x)^m)$$ and $$k[t]/(t^m)$$ have characteristic $$p$$ for all $$m \geq 1$$, so there's nothing unusual about them winding up as isomorphic to each other for all $$m$$.) I will describe two methods, the first one being more concrete.

Method 1: The elements of $$k[t]/(t^m)$$ are uniquely expressible as $$c_0 + c_1t + \cdots + c_{m-1}t^{m-1} \bmod t^m$$ with $$c_j \in k$$. Inside $$\mathbf F_p[x]/(Q^m)$$, the elements are uniquely expressible as $$a_0(x) + a_1(x)Q(x) + \cdots + a_{m-1}(x)Q(x)^{m-1}$$ with $$a_j(x) = 0$$ or $$\deg(a_j(x)) < \deg Q$$. But this way of writing the elements of $$\mathbf F_p[x]/(Q^m)$$ is terrible in order to set up a ring isomorphism with $$k[t]/(t^m)$$ since those base $$Q$$ digits $$a_j(x)$$ are not multiplicatively closed (in contrast to the $$c_j$$ in $$k$$). What we need to do is find a copy of the field $$k$$ of order $$p^f$$ inside $$\mathbf F_p[x]/(Q^m)$$.

The polynomial $$t^{p^f} - t$$ splits completely over the field $$\mathbf F_p[x]/(Q)$$, so for each $$m \geq 1$$ the $$Q$$-adic Hensel's lemma tells us we can lift each of those roots mod $$Q$$ uniquely to a root of $$t^{p^f}-t$$ in $$\mathbf F_p[x]/(Q^m)$$. Set $$k_m := \{b(x) \bmod Q^m : b(x)^{p^f} \equiv b(x) \bmod Q^m\},$$ which is the roots of $$t^{p^f}-t$$ in $$\mathbf F_p[x]/(Q^m)$$. This set is closed under addition and multiplication and each nonzero element is invertible: if $$b(x) \not\equiv 0 \bmod Q^m$$ then $$b(x) \not\equiv 0 \bmod Q$$ (the only lift of $$0 \bmod Q$$ as a root is $$0 \bmod Q^m$$), so $$\gcd(b(x),Q^m) = 1$$. So $$k_m$$ is a field of order $$p^f$$ inside $$\mathbf F_p[x]/(Q^m)$$ and in fact it's the only such field inside $$\mathbf F_p[x]/(Q^m)$$ thanks to the unique lifting of roots of $$t^{p^f}-t$$ from modulus $$Q$$ to modulus $$Q^m$$.

Since $$k_m$$ inside $$\mathbf F_p[x]/(Q^m)$$ is a set of representatives for $$\mathbf F_p[x]/(Q)$$, we can write every element of $$\mathbf F_p[x]/(Q^m)$$ uniquely as $$b_0(x) + b_1(x)Q(x) + \cdots + b_{m-1}(x)Q(x)^{m-1} \mod Q(x)^m$$ where $$b_j(x) \bmod Q(x)^m \in k_m$$. This way of writing the elements of $$\mathbf F_p[x]/(Q^m)$$ looks just like the usual way of writing the elements of $$k[t]/(t^m)$$ and it shows $$k_m[Q \bmod Q^m]$$ fills up $$\mathbf F_p[x]/(Q^m)$$

Since $$k$$ and $$k_m$$ are fields of equal size $$p^f$$ there is a field isomorphism $$\varphi_m \colon k \to k_m$$ (in fact there are $$m$$ isomorphisms, but that doesn't matter). Extend $$\varphi_m$$ to a ring homomorphism $$k[t] \to \mathbf F_p[x]/(Q^m)$$ by mapping $$t$$ to $$Q \bmod Q^m$$: $$\sum_{i} c_it^i \mapsto \sum_i \varphi_m(c_i)Q^i \bmod Q^m.$$ This is surjective since $$\mathbf F_p[x]/(Q^m)$$ is generated as a ring by $$k_m$$ and $$Q \bmod Q^m$$. Since $$t^m \mapsto Q^m \bmod Q^m = 0$$, we get an induced surjective ring homomorphism $$k[t]/(t^m) \to \mathbf F_p[x]/(Q^m)$$. Both have order $$p^{fm}$$, so this is a ring isomorphism.

Method 2: Since $$Q$$ is irreducible in $$\mathbf F_p[x]$$, the ring $$\mathbf F_p[x]/(Q^m)$$ is unchanged up to isomorphism if we replace $$\mathbf F_p[x]$$ with its $$Q$$-adic completion $$\mathbf F_p[x]_Q$$: $$\mathbf F_p[x]/(Q^m) \cong \mathbf F_p[x]_Q/(Q^m)$$ as rings. The residue field of $$\mathbf F_p[x]_Q$$ is isomorphic to $$\mathbf F_p[x]/(Q)$$, which is isomorphic to $$k$$. The completion $$\mathbf F_p[x]_Q$$ is the ring of integers of the $$Q$$-adic field completion $$\mathbf F_p(x)_Q$$, and the structure of local fields of positive characteristic says they are all isomorphic to the formal Laurent series field over the residue field. Thus $$\mathbf F_p(x)_Q \cong k((t))$$ as valued fields, so their rings of integers are isomorphic: $$\mathbf F_p[x]_Q \cong k[[t]]$$. The isomorphism identifies powers of the maximal ideal on both sides, and the maximal ideal of $$\mathbf F_p[x]_Q$$ is $$(Q)$$ since $$Q$$ is a uniformizer in the $$Q$$-adic completion. Thus $$\mathbf F_p[x]_Q/(Q^m) \cong k[[t]]/(t^m)$$ for each $$m \geq 1$$. Both sides simplify to quotient rings of $$\mathbf F_p[x]$$ and $$k[t]$$, just as $$\mathbf Z_p/(p^m) \cong \mathbf Z/(p^m)$$: $$\mathbf F_p[x]/(Q^m) \cong k[t]/(t^m).$$

Here is an illustration of Method 1.

Example. Let $$Q(x) = x^2 - 2$$ in $$\mathbf F_5[x]$$. Then $$\mathbf F_5[x]/(Q^3) \cong \mathbf F_{25}[t]/(t^3)$$. To write down an explicit ring isomorphism we need a copy of $$\mathbf F_{25}$$ in $$\mathbf F_5[x]/(Q^3)$$. We can view $$\mathbf F_{25}$$ as $$\mathbf F_5(\alpha)$$ where $$\alpha^2 = 2$$. One root of $$t^2 - 2$$ in $$\mathbf F_5[x]/(x^2-2)$$ is $$x \bmod x^2-2$$, and the unique lifting of this to a root of $$t^2 - 2$$ in $$\mathbf F_5[x]/(Q^3)$$ is $$r = x + xQ + 4xQ^2 = x + x(x^2-2) + 4x(x^2-2)^2 \bmod Q^3$$ by an explicit search. Therefore the copy of $$\mathbf F_{25}$$ inside $$\mathbf F_5[x]/(Q^3)$$ is $$\mathbf F = \mathbf F_5 + \mathbf F_5r$$ and $$\mathbf F_5[x]/(Q^3) = \mathbf F + \mathbf F Q + \mathbf F Q^2 \bmod Q^3.$$ The right side naturally looks like $$\mathbf F_{25}[t]/(t^3)$$ and we get an isomorphism $$\mathbf F_{25}[t]/(t^3) \to \mathbf F_5[x]/(Q^3)$$ once we decide on an isomorphism between $$\mathbf F_{25}$$ and $$\mathbf F$$.

• That's very interesting. To show that $\mathbb F_p[x]/(Q(x)^m)$ (where $Q$ is irreducible) contains a field isomorphic to $k=\mathbb F_p[x]/(Q(x)),$ I argued as follows: Replace $m$ with $q\ge m$ where $q$ is a power of $p.$ Given the required field for $q,$ we can then take its image under $\mathbb F_p[x]/(Q(x))^q\rightarrow \mathbb F_p[x]/(Q(x))^m.$ But $(Q(x))^q=(Q(x^q))$ and $(Q(x))^q\cap \mathbb F_p [x^q]=Q(x^q)\mathbb F_p [x^q].$ Hence $\mathbb F_p[x]/(Q(x))^q$ contains $\mathbb F_p[x^q](Q(x))^q/(Q(x))^q$ which is isomorphic to $\mathbb F_p[x^q]/Q(x^q)\mathbb F_p[x^q],$ i.e to $k.$ Commented Mar 5, 2023 at 19:48
• I think that can be seen in your example-the approach I mention would give $x^5$ mod $Q^3$ in place of $r,$ but indeed $r=-x^5$ mod $Q^3.$ Commented Mar 5, 2023 at 19:53
• It looks like the end of your second comment has a typographical error since what you wrote as $\mathbf F_p[x^q](Q(x))^q/(Q(x))^q$ is not a ring. Commented Mar 5, 2023 at 20:02
• You're right of course, it should say $(\mathbb F_p[x^q]+(Q(x))^q)/(Q(x))^q.$ Commented Mar 5, 2023 at 20:08
• Ah, that's a nice alternative way to find the field of order $p^f$ inside $\mathbf F_p[x]/(Q^m)$. Commented Mar 5, 2023 at 20:24