A few past exam questions

							     A few past exam questions
                                                                √
    Consider the element 10 + 2α inside the ring Z[α] for α = −10. Let’s try to factorize it into
irreducible elements rather than maximal ideals. For this as well, computing the norm is convenient.
One obvious factorization stares us in the face:

                                          10 + 2α = α(−α + 2)

If we compute norms, we get

                              N (α) = 10 = 2 × 5, N (−α + 2) = 14 = 2 × 7

So if either α or −α + 2 were reducible, then some element of Z[α] would have norm 2. But clearly,

                                      2 = N (n + mα) = n2 + 10m2

has no integral solutions in n, m. So α and −α + 2 are irreducible. Therefore, we’ve found a decom-
position of 10 + 2α into irreducible elements.
    To compute the decomposition into maximal ideals, we need to compute the decompositions of
(2), (5), and (7) in Z[α]. Since x2 + 10 ≡ x2 mod 2, we get
                                                       2
                                                (2) = P2

for P2 = (2, α). Similarly,
                                                       2
                                                (5) = P5
for P5 = (5, α). Finally,

                            x2 + 10 ≡ x2 + 3 ≡ x2 − 4 = (x − 2)(x + 2)   mod 7

so
                                                7 = P7 P7
where P7 = (7, α − 2) and P7 = (7, α + 2). From this computation, it is clear that
                                                        2
                                           (10 + 2α) = P2 P5 P7

or
                                                        2
                                           (10 + 2α) = P2 P5 P7
To find out which is correct, we need to see if 10 + 2α belongs to P7 or P7 . But

                                        10 + 2α = 2 × 7 + 2(α − 2)

so that 10 + 2α ∈ P7 . Hence,
                                                        2
                                           (10 + 2α) = P2 P5 P7
is the correct decomposition. You might note at this point with little difficulty that this relates to the
previous decomposition via the formula
                                            (α) = P2 P5
and
                                             (α − 2) = P2 P7
   Exercise: Use these computations to come up with another decomposition of 10+2α into irreducible
                      √
elements, revealing Z[ −10] not to be a unique factorization domain.
   Now we consider the case of Z[α] where α is a root of x3 + 9x + 3. It is straightforward to check
that Z[α] = OK for K = Q[α]. The point is that the discriminant of the basis {1, α, α2 } is 35 × 17 × 19
and you need to use a theorem about Eisenstein polynomials.

                                                    1
    One problem from a previous exam concerns finding ideals in OK with norm 49. Of course, such
an ideal must have prime ideal factors all dividing (7). So we need to consider the decomposition of
(7) inside OK . We have

                        x3 + 9x + 3 ≡ x3 + 2x + 3 = (x + 1)(x2 − x + 3)   mod 7

and a quick check will show that x2 − x + 3 is irreducible in F7 [x]. Therefore,

                                             (7) = P7 P49

where
                                            P7 = (7, α + 1)
and
                                         P49 = (7, α2 − α + 3)
                2
So we see that P7 and P49 are the ideals in question. To check for principality, the first thing to do
is to compute norms of some elements. We see that

                         N (α + 1) = −N (−1 − α) = −[(−1)3 + 9(−1) + 3] = 7,

that is, the element α + 1 ∈ P7 has norm equal to N (P7 ). Therefore, (α + 1) = P7 and P7 is principal.
              2
A fortiori, P7 is principal. But then, since
                                                   −1
                                            P49 = P7 (7),

it is also principal.




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