By Tom Weston

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**Extra info for Algebraic Number Theory [Lecture notes]**

**Example text**

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. Note that the definition says nothing about monic polynomials in R[x] actually having roots in K; it says only that if they do have roots in K, then these roots lie in R. Note also that there is nothing about the minimal polynomial of α in the definition; any monic polynomial at all will do, irreducible or not.

There are three possibilities for the factorization of f¯(x) in Fp [x]. First of all, f¯(x) could be irreducible. Second, f¯(x) could factor as a product of distinct, monic linear (and therefore irreducible) polynomials. Third, f¯(x) could factor as the square of a single monic linear polynomial. We will consider all three cases separately. Suppose first that f¯(x) is irreducible in Fp [x]. Then Fp [x]/(f¯(x)) is a field, so OK /pOK is as well. pOK is therefore a prime ideal of OK , by the definition of prime ideal, so it does not factor any further.

The problem with R is that it is missing certain elements; we will see the full solution in the next section. 2. 1. Integrally closed rings. The key to our search for the right special subring of a number field K is the “good factorization theory” condition. 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].