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I have been working through a book on Ring/Algberaic number theory recently and the book only has solutions for even questions. I was wondering if somebody could help with a solution for the following question.

I am so far unsure, the route I have gone down has included using the minimal polynomial x^3 - 6. Also then reducing mod p, yet I cannot create a full solution.

Thanks in advance

Bernard
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RedPvwer
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  • Which book is it? – lhf Feb 28 '18 at 23:49
  • reduce mod p and find a root in Z/pZ. That root will allow you to factorize. Example: p = 5 then x = 1 is a root so write x^3 - 6 as (x-1)(x^2-x+1). Pass to ideals and write (5) = (5,a-1)(5,a^2-a+1) where a = cbrt(6) – TCiur Feb 28 '18 at 23:50
  • Notes by loeffer, a former lecturer of mine – RedPvwer Feb 28 '18 at 23:51
  • @TCiur I have reached similar points, yet after this would I then find the roots of the quadratic and the product of the ideals generated by the roots would be the prime factorisation? – RedPvwer Feb 28 '18 at 23:53
  • in this case the quadratic has no roots, and so in particular O[x]/(5,x^2-x+1) is an integral domain, so (5,x^2-x+1) is a prime ideal. – TCiur Feb 28 '18 at 23:57

1 Answers1

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Sorry for this scrappy answer.

The idea is to exploit the isomorphism:

$Z[a] \rightarrow Z[x]/(f(x))$

Where $a$ is an algebraic integer and $f()$ is its minimal polynomial. $a$ is sent to $x$ and vice versa.

You should know that prime ideals $\mathfrak{p}$ in $Z[a]$ correspond to integral domains $Z[a]/\mathfrak{p}$. The second idea is to find zero divisors of $Z[a]/\mathfrak{p}$ in order to decompose $Z[a]/\mathfrak{p}$ and hence factor $\mathfrak{p}$. This is difficult and so we use the first idea to map the problem to the polynomial side:

$Z[a]/\mathfrak{p}(a) \rightarrow (Z[x]/(f(x))) / \mathfrak{p}(x) = Z[x]/(f(x),\mathfrak{p}(x))$

*where $\mathfrak{p}(a) \rightarrow \mathfrak{p}(x)$ means we are swapping all instances of $a$ for $x$

Now we look for zero divisors in $Z[x]/(f(x),\mathfrak{p})$ but these are just roots of $f(x)$ mod $\mathfrak{p}$.

Example for $\mathfrak{p} = (5)$, $a = 6^{1/3}$:

$Z[a]/(5) \rightarrow (Z[x]/(x^3 - 6)) / (5) = Z[x]/(x^3-6,5)$

This makes problem easier. Factor $x^3-6$ in $Z[x]/(5)$ as $(x-1)(x^2-x+1)$ and note:

$Z[x]/(x^3-6,5) = Z[x]/(x-1,5) \times Z[x]/(x^2-x+1,5)$

Map back to the "number field" side and then consider the ideals:

$(a^3-6,5) = (a-1,5)(a^2-a+1,5)$

TCiur
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