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When I was reading an article about linear recurrence relations, I saw this notation:

$$P=\mathfrak{p}_1^{e_1}\mathfrak{p}_2^{e_2}...\mathfrak{p}_r^{e_r}$$ $$ N(\mathfrak{p}_{i})=P^{k_i}$$

What is the meaning of the notation $ N(\mathfrak{p}_{i})$? and also what is $k_i$ (degrees of prime functions (mod P))?

Here is some parts of the article:

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SMA.D
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This is a theorem from algebraic number theory. It is about how the prime ideal $(p)$ splits in the extension field $F=\Bbb{Q}(\theta)$.

The letter $N$ stands for norm (in the sense of ideals). $N(\mathfrak{p}_i)$ is the index of the prime ideal $\mathfrak{p}_i$ in the ring of integers $\mathcal{O}_F$ of the field $F$. It is a power of $p$, because the quotient ring $\mathcal{O}_F/\mathfrak{p_i}$ is a finite field of characteristic $p$.

For example if $F(x)=x^3-2$ we get $F=\Bbb{Q}(\root3\of2)$. It is not too difficult to see that in this case $\mathcal{O}_F=\Bbb{Z}[\root3\of2]$, but not too easy either, so I skip that part.

If we study the splitting of the prime $p=5$ we look at the factorization of $F(x)$ modulo $5$. We easily find that $$ x^3-2\equiv(x-3)(x^2-2x-1) \pmod5, $$ but here the factor $x^2-2x-1$ has no zeros modulo five, so it is irreducible. Anyway the result states that the ideal $(5)=5\mathcal{O}_F$ factors as $$ (5)=(5,\root3\of2-3)(5,\root3\of4-2\root3\of2-1). $$ You see that the other generator of those ideals was gotten by plugging in $\root3\of2$ in place of $x$ in the modulo $5$ factorization of $F(x)$.

The result is due to Dedekind. It does not hold always, but it can be shown to hold in two important cases: 1) There are no repeated factors, i.e. $e_i=1$ for all $i$. 2) The powers of $\theta$ form a basis of $\mathcal{O}_F$ as a free abelian group.

The integers $k_i$ (aka inertia degrees) measure the size of the quotient rings - the ring $\mathcal{O}_F/(p,\phi_i(\theta))$ has $p^{k_i}$ elements. They are also the degrees of the factors $\phi_i(x)$. So in the above example $\Bbb{Z}[\root3\of2]/(5,\root3\of2-3)$ is a field of $5^1$ elements and $\Bbb{Z}[\root3\of2]/(5,\root3\of4-2\root3\of2-1)$ is a field of $5^2$ elements reflecting the facts that the mod $5$ factors $x-3$ and $x^2-2x-1$ have degree $1$ and $2$ respectively.

Not sure if this helps. I should not attempt to reproduce scores of pages from an introduction to the theory of algebraic numbers :-/

Jyrki Lahtonen
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  • Thank you for answering. Is there a more simple way to find $k_i$'s without finding factorization of $F(x) (mod p)$, Or an algorithm to find factorization of $F(x) (mod p)$ easily? – SMA.D Jul 25 '15 at 17:20
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    Judging from the fact Mathematica does that unbelievable well factoring is not too difficult. Berlekamp's algorithm is quite good, and Jack's answer there hints at something better. That question is mostly about finding linear factors, but IIRC it generalizes. – Jyrki Lahtonen Jul 25 '15 at 17:28