7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ That is not true. $\endgroup$ Mar 5, 2023 at 12:56
  • $\begingroup$ Are you asking for an isomorphism as vector spaces over $\mathbb{Z}/p$, as rings, or in some other sense? $\endgroup$ Mar 5, 2023 at 13:06

1 Answer 1

15
$\begingroup$

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 $\mathbf F$.

$\endgroup$
6
  • $\begingroup$ 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 $k.$ $\endgroup$
    – Tom
    Mar 5, 2023 at 19:48
  • $\begingroup$ 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 $Q^3.$ $\endgroup$
    – Tom
    Mar 5, 2023 at 19:53
  • $\begingroup$ 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. $\endgroup$
    – KConrad
    Mar 5, 2023 at 20:02
  • $\begingroup$ You're right of course, it should say $(\mathbb F_p[x^q]+(Q(x))^q)/(Q(x))^q.$ $\endgroup$
    – Tom
    Mar 5, 2023 at 20:08
  • $\begingroup$ Ah, that's a nice alternative way to find the field of order $p^f$ inside $\mathbf F_p[x]/(Q^m)$. $\endgroup$
    – KConrad
    Mar 5, 2023 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.