By Tom Weston

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As we have seen, it is unreasonable to expect unique factorization, although there is still some hope that we may be able to get unique factorization of ideals. What we need, then, is some condition which √ is weaker than UFD but still strong enough to eliminate the problem case of Z[ 3]. The correct condition turns out to be the following. 1. Let R be an integral domain contained in some field K. An element α ∈ K is said to be integral over R if it satisfies some monic polynomial in R[x]. R is said to be integrally closed in K if every element in K which is integral over R actually lies in R.

A positive rational prime p can be written as x2 + y 2 with x, y ∈ Z if and only if p = 2 or p ≡ 1 (mod 4). Furthermore, this decomposition is unique up to switching x and y and negating either (or both) x or y. Proof. 7. In fact, uniqueness also follows easily, since there are exactly 8 primes x + yi dividing any p ≡ 1 (mod 4) (two conjugates times four units) and these all have x and y the same up to negation and switching the factors. 4. Failure of unique factorization. Having given one example where everything works perfectly, let us now give several where things do not work.

Since d is squarefree this implies that b ∈ Z; thus we do not get any additional integers in this case. The other case is that a = a1 /2, where a1 ∈ Z is odd. Since a21 − b2 d ∈ Z 4 we must have b = b1 /2 where b1 in Z is also odd. Substituting this in, we find that a21 − b21 d ≡ 0 (mod 4), this being an ordinary congruence over the integers. Now, since a1 and b1 are both odd, a21 ≡ b21 ≡ 1 (mod 4). Substituting these in, we find that 1 − d ≡ a21 − b21 d ≡ 0 (mod 4), 2. ALGEBRAIC INTEGERS 41 so d≡1 (mod 4).