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Definition: Let $f(x)$ be an irreducible monic polynomial of degree $d$ with integer coefficients, and suppose $f(r) \equiv 0 \bmod p$, where $p$ is prime. Let $g_r$ be a polynomial of degree $e$ $(0

Questions: (1) It it possible to predict without searching which prime / root combinations have representations (preferably for d > 2)? (2) Some $f$ appear to have only a finite number of prime / root combinations without representations e.g. the above example. Is it possible to predict these?

Some pointers to relevant theory would be much appreciated.

The questions arise from looking at the number field sieve, where norms of $\prod_{}{}(a+b\alpha)$ are products of particular primes. If the norm of such a product is divisible by some prime $p$, $a+br \equiv 0 \bmod p$, where $r$ is one of the roots of $f \equiv 0 \bmod p$. I am trying to understand Lenstra and Lenstra's 'The development of the number field sieve', but have difficulty translating representations into terms of ideals.

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    It seems that you're asking whether the prime ideal $(p,\alpha-r)$ of $\mathbb Z[\alpha]$ is principal, so if you want a (rather weak) pointer, search for "how to tell whether an ideal is principal", "what is the density of principal ideals" etc. (sorry if it's not helpful). – 2017-02-06
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    Thanks. I don't think I am asking whether the prime ideal $(p,\alpha-r)$ is principal, assuming you mean that $\rho(p_r)$ is the generator. I tried this for $\rho(3_2)$, and the two ideals are different. I have added a bit more explanation to the original post which may clarify things a bit. – 2017-02-07
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    my reasoning was this: you want the ideal $(g_r(\alpha))$ to have norm $p$ (here I already forgot about the sign), and such an ideal is necessarily of the from $(p,\alpha-s)$ for some $s\in\mathbb Z$ such that $f(s)\equiv 0$ mod $p$; it felt like $s$ should be $r$, but I didn't check it – 2017-02-07
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    Apologies - I had a bug in my program. The two ideals $(\alpha+1),(3,\alpha-2)$ are identical, as proved by $(\alpha+1)(-\alpha^2-\alpha-3)=\alpha-2$, $(\alpha+1)(\alpha^2+\alpha-4)=3$, and $(\alpha-2)(-\alpha^2+2\alpha)+3(-2\alpha^2)=\alpha+1$. Many thanks for your help. I will follow up on your suggested searches. – 2017-02-08

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