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MATH 254A: NUMBER FIELDS AND RINGS OF INTEGERS
BRIAN OSSERMAN
In this lecture we will introduce rings of integers, and develop some of their basic
properties. We will also introduce the norm, trace, and discriminant.
1. Rings of integers
The first definition of algebraic number theory is the following:
Definition 1.1. A number field K is an extension of Q of finite degree.
What is the best way to generalize rings such as Z[i]? Although the definition
may appear unmotivated at first, we will mainly consider the following rings:
Definition 1.2. Given a number field K, the ring of integers OK of K is defined
to be the set of elements α ∈ K such that α satisfies a monic polynomial with
coefficients in Z; equivalently, there exists f(x) ∈ Z[x] monic and such that f(α) =
0.
The ring of integers of a number field is intended to generalize the ring Z inside
of Q. Thus, we ought to check that Z is in fact the ring of integers of Q.
Definition 1.3. We say that an integral domain R is a unique factorization domain, or UFD, if every non-zero element may be factored uniquely as a product of
irreducible elements, up to multiplication by units. In particular, every irreducible
element is prime.
Lemma 1.4. Let R be a unique factorization domain. Then R is integrally closed
in its field of fractions.
Proof. Given α = r/s in the field of fractions of R, we may assume that r, s have
no non-unit common factors. Suppose that α satisfies a monic polynomial xm +
am−1 xm−1 + · · · + a1 x + a0 , with ai ∈ R for all i. Because R is a UFD, if r/s 6∈ R,
there is some prime p such that p|s, whence p 6 |r. But substituting r/s for x in
the polynomial, and multiplying through by sm , we find r m + am−1 r m−1 s + · · · +
a1 rsm−1 + sm = 0, so p|r m , and hence p|r, which is a contradiction.
Corollary 1.5. Z is integrally closed, hence is the ring of integers of Q.
√
Exercise 1.6. Check that for n ∈ N square-free, the ring of integers of Q( n) is
given as {a + bω : a, b ∈ Z}, and where ω is given as follows:
√
(I) if n 6≡ 1 (mod 4), then ω = n;
√
(II) if n ≡ 1 (mod 4), then ω = 1+2 n .
Proposition 1.7. A ring of integers OK is an integral domain (in particular, a
ring).
More generally, if R is an integral domain, and L a field containing R, then the
set S of elements of L which are integral over R is an integral domain.
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BRIAN OSSERMAN
The proof of the proposition is very similar to the proof that if a field is extended
by a finite number of algebraic elements, the resulting field extension is algebraic.
In fact, the statement for fields follows as a special case.
Lemma 1.8. Let R be an integral domain, L any field containing R, and W ⊂ L
a non-zero finitely-generated R-module. Given any α ∈ L, suppose that αW ⊂ W .
Then α is integral over R; that is, it satisfies a monic polynomial with coefficients
in R. Conversely, if α is integral over R, then R[α] is a W as above.
In particular, R[α] is a finitely-generated R-module if and only if α is integral
over R.
Proof. Let mα : W → W denote the map given by multiplication by α, which we
observe is R-linear. Given a set of generators (w1 , . . . , wn ) for W over R, we find
that mα is represented by an P
n × n matrix (mi,j )i,j with coefficients in R. Now,
n
inside of L, we have αw1 =
i=1 mi1 wi , by definition. But it follows that the
n
n
map αIn − (mi,j )i,j : L → L has w1 in its kernel, hence the determinant is 0.
Expanding the determinant, we obtain a monic polynomial for α with coefficients
in R.
Conversely, if f(α) = 0 for f(x) ∈ R[x] monic of degree d, it is clear that
1, α, . . . , αd−1 generates R[α] as a R-module.
Proof of Proposition. Since S is contained in the field L, it is clear that if it is a
ring, it is an integral domain. It also is clear that if α is integral, then so is −α, since
we can just alternately change the signs in any monic integral polynomial satisfied
by α. It therefore suffices to check that if α1 , α2 are integral over R, then so are
α1 +α2 and α1 α2 . Now, if α1 , α2 satisfy monic polynomials in R[x] of degrees d1 , d2
respectively, consider the module W = R < (αi1 αj2 )06i<d1 ,06j<d2 >. One checks
that because of the monic polynomials satisfied by the αi , we have αi W ⊂ W for
i = 1, 2, and hence α1 α2 W ⊂ W and (α1 + α2 )W ⊂ W , so the proposition follows
from the lemma.
2. Norm, trace, and discriminant
We begin with some basic definitions relating to an arbitrary field extension L/K
of some finite degree n. Given any α ∈ L, we denote by mα : L → L the map given
by multiplication by α, which we observe is K-linear automorphism of the K-vector
space L.
Definition 2.1. Given L/K and α ∈ L, the norm NL/K (α) is given by det mα .
The trace TrL/K (α) is given by Tr mα . Given also α1 , . . . , αn ∈ L, the discriminant DL/K ((αi )i ) of {αi }i is given by det(TrL/K (αi αj ))i,j .
These concepts are all very useful in dealing with concrete calculations in algebraic number theory.
Observation 2.2. The trace is additive and K-linear, and the norm is multiplicative.
Now suppose that L/K is separable, as will always be the case in characteristic
0, so in particular for number fields. Fix an algebraic closure K̄, and let σ1 , . . . , σn
be the distinct imbeddings of L into K̄ fixing K (if L/K is Galois, these are all
related by automorphism of L). We have the following:
MATH 254A: NUMBER FIELDS AND RINGS OF INTEGERS
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Proposition 2.3. With L/K separable, we have
TrL/K (α) =
n
X
σi (α),
i=1
NL/K (α) =
n
Y
σi (α),
i=1
and
DL/K ((αi )i ) = (det(σi (αj ))i,j )2 .
Exercise 2.4. Prove the first two parts of the proposition in the following steps:
(1) Show that if α ∈ L, and f(x) ∈ K(x) is the monic minimal polynomial for
α, with degree d, then det(xI − mα ) = f(x)n/d .
(2) Using the fact that if E/F is a separable field extension of finite degree,
then any imbedding E → F̄ has exactly [F : E] extensions to imbeddings
Qn
F → F̄ , show that in the above notation, f(x)n/d = i=1 (x − σi (α))
(3) Conclude the statements on the norm and trace by comparing with the
appropriate coefficients of det(xI − mα ).
Proof. Only the discriminant assertion remains. By the trace formula, we have
TrL/K (αi αj ) =
n
X
σℓ(αi )σℓ (αj ).
ℓ=1
It follows that we obtain the matrix identity
(TrL/K (αi αj ))i,j = (σℓ (αi ))i,ℓ (σℓ(αj ))ℓ,j ,
and taking the determinant of both sides, and using that the determinant is invariant under matrix transposition, we get the desired identity.
Corollary 2.5. In the situation of the proposition, suppose that α and the αi are
all integral over some ring R ⊂ K. Then TrL/K (α), NL/K (α), and DL/K ((αi )i )
are integral over R.
Proof. First note that if ω ∈ L is integral over R, then σi (ω) is still integral over
R for any i, and in fact satisfies the same monic integral polynomial, since by
hypothesis applying σ fixes the coefficients, as they are in R ⊂ K. Since the trace,
norm and discriminant are obtained as sums and products of such elements, by
Proposition 1.7 we obtain integrality over R.
Lemma 2.6. Given
Pn α1 , . . . , αn ∈ L, and an n × n matrix (ai,j )i,j with coefficients
in K, let α′i = j=1 ai,j αj . Then DL/K ((α′i )i ) = (det(ai,j )i,j )2 DL/K ((αi )i ).
Proof. Observe that because ai,j ∈ K for all i, j, we have
(σj (α′i ))i,j = (ai,ℓ )i,ℓ(σj (αℓ))ℓ,j .
Then taking determinants and squaring yields the desired identity.
Lemma 2.7. If L/K is separable, and we are given α1 , . . . , αn ∈ L, we have
DL/K ((αi )i ) 6= 0 if and only if (αi )i forms a basis for L over K.
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BRIAN OSSERMAN
Proof. Because L is separable over K, by the primitive element theorem we can
set L = K(α) for some α ∈ L. Using the basis (1, α, . . . , αn−1) for L over K, and
the formula for the discriminant of the above proposition,
Q we find it is given by
the square of the Vandermonde determinant, which is i>j (σi (α) − σj (α)) 6= 0.
The fact that the discriminant is non-zero for any basis of L then follows from the
formula of the previous lemma.
Conversely, any K-linear dependence of the αi gives the same K-linear dependence of the σj (αi ) for any fixed j, so we obtain a linear dependence on the columns
of the matrix (σj (αi ))j,i , which then has determinant 0.