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I have been using the database of lmfdb.org to find the integral basis of a number field. Now, I want to utilize PARI/GP in multiplying algebraic integers. However, I have encountered a problem. PARI/GP uses the integral basis "nf.zk" in its computations, which apparently is not always the same as the "nfbasis(f)", which is the integral basis that lmfdb.org provides.

For example, we have the following code from PARI/GP:

? f = x^3 - x^2 + 2*x + 8
nf = nfinit(f)
nf.zk
%1 = [1, x, 1/2*x^2 - 1/2*x + 1]
? nfbasis(f)
%2 = [1, x, 1/2*x^2 - 1/2*x]

Now, my questions are:

  1. Why are nf.zk and nfbasis(f) different?
  2. Why does PARI/GP use nf.zk instead of nfbasis(f)?
  3. Lastly, can I tell PARI/GP to use nfbasis(f) instead of nf.zk?
Charles
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Butterfly
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1 Answers1

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  1. When we take the trouble to initialize an nf structure with nfinit, we perform precomputations to speed up later work. Here, nfinit first computes the integer basis by calling nfbasis, which returns the (canonical) HNF basis, then LLL-reduces it with respect to the T2 norm. The LLL-reduced basis is usually different from the HNF one, but it usually has smaller elements.

  2. This LLL reduction can be expensive (in particular when the degree is large) but it ensures that time complexities are bounded in terms of the field discriminant instead of the size of the input polynomial. I believe all polynomials defining number fields in the lmfdb were run through polredabs which ensures their coefficients are small (in terms of the field discriminant), but the HNF integer basis may still be much larger than the LLL one. Additionally, if an algebraic integer has small T2 norm, its expression in terms of the LLL-reduced basis is guaranteed to have small coefficients, whereas it can have much larger coefficients on the HNF basis.

  3. In pari-2.14 (which is not released yet but available via git or through nightly snapshots on the PARI/GP website), you can use nfinit(, 4), which removes the LLL reduction step. This speeds up the initialization, but usually slows down every operation involving the resulting nf.

? f = x^3 - x^2 + 2*x + 8
? nfinit(f,4).zk
%2 = [1, x, 1/2*x^2 - 1/2*x]
K.B.
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