Timeline for Quotients of number fields by certain prime powers
Current License: CC BY-SA 4.0
11 events
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| Mar 5, 2023 at 20:35 | comment | added | KConrad | A way of estimating the solution $\alpha$ of $t^2 - 2 = 0$ in the completion $\mathbf F_5[x]_{x^2-2}$ that is congruent to $x \bmod x^2-2$ is to recognize that in this completion, the solution of $t^2-2=0$ is the Teichmueller representative of $x$. In a local field with prime $\pi$ and residue field of order $q$, each unit $u$ with Teichmueller representative $\omega$ has $\omega = \lim_{n \to \infty} u^{q^n}$. Since $u^{q^n} \equiv u^{q^{n-1}} \bmod \pi^n$ for all $n \geq 1$, $\omega \equiv u^{q^{m-1}}\bmod \pi^m$. Thus $\alpha \equiv x^{25^{m-1}} \bmod (x^2-2)^m$ | |
| Mar 5, 2023 at 20:24 | comment | added | KConrad | Ah, that's a nice alternative way to find the field of order $p^f$ inside $\mathbf F_p[x]/(Q^m)$. | |
| Mar 5, 2023 at 20:08 | comment | added | Tom WIlde | You're right of course, it should say $(\mathbb F_p[x^q]+(Q(x))^q)/(Q(x))^q.$ | |
| Mar 5, 2023 at 20:02 | comment | added | KConrad | 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. | |
| Mar 5, 2023 at 19:53 | comment | added | Tom WIlde | 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 $Q^3.$ | |
| Mar 5, 2023 at 19:48 | comment | added | Tom WIlde | 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 $k.$ | |
| Mar 5, 2023 at 19:25 | history | edited | KConrad | CC BY-SA 4.0 |
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| Mar 5, 2023 at 19:13 | history | edited | KConrad | CC BY-SA 4.0 |
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| Mar 5, 2023 at 18:26 | vote | accept | Tom WIlde | ||
| Mar 5, 2023 at 17:10 | history | edited | KConrad | CC BY-SA 4.0 |
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| Mar 5, 2023 at 17:04 | history | answered | KConrad | CC BY-SA 4.0 |