Commutative algebra - Department of Mathematical Sciences - old ...

1.8. FIELDS 19
1.8. Fields
1.8.1. Definition. Let p ∈ Z be a prime number. The factor ring Fp = Z/(p) is a
field with p elements. Together with Q they constitute the prime fields.
1.8.2. Proposition. Let K be a field then the polynomial ring K[X] is a principal
ideal domain.
Proof. Let d = 0 be a polynomial of lowest degree in an ideal I. Given f ∈ I then
by 1.6.6 f = qd + r with r = f − qd ∈ I. By degree considerations r = 0 and
I = (d).
1.8.3. Corollary. Let K be a field then the polynomial ring K[X] is a unique
factorization domain.
Proof. Follows from 1.5.6.
1.8.4. Definition. A subfield is a subring, which is a field. A field extension is the
inclusion of a subfield K ⊂ L in a field. A finite field extension K ⊂ L is an
extension, where L is finite dimensional as a vector space over K.
1.8.5. Example. (1) Let K be a field and f an irreducible polynomial in K[X].
Then K ⊂ K[X]/(f) is a finite field extension.
(2) Let K ⊂ R ⊂ L be a subring in a finite field extension. Then R is a field.
Namely multiplication on R with a nonzero element of R is a K-linear map
on the finite dimensional K-vector space R and therefore an isomorphism.
1.8.6. Proposition. (1) Let K be a field and f a polynomial in K[X]. Then there
is a finite field extension K ⊂ L such that f factors in linear factors in L[X].
(2) If K ⊂ L1 and K ⊂ L2 are finite field extensions then there is a finite field
extension K ⊂ L such that L1 ∪ L2 ⊂ L.
Proof. (1) Assume f irreducible. In L = K[X]/(f) the class X + (f) is a root of
f. In general proceed adjoining roots of irreducible factors of f. (2) An element x
in a finite field extension K ⊂ K ′ is the root of the irreducible monic polynomial
f generating the kernel of the evaluation homomorphism K[X] → K ′ , X ↦→ x.
Now proceed by (1) adjoining elements in L2 to L1.
1.8.7. Proposition. Let p ∈ Z be a prime number. For any power q = p n there is
a field Fq with q elements, unique up to isomorphism.
Proof. Let Fp ⊂ K be a field extension, where X q − X factors into linear factors,
1.8.6 (1). The subset of roots is the set of elements fixed under n-times the
Frobenius and therefore a subring being a subfield by 1.8.5 (2). The derivative
(X q − X) ′ = −1 so by 1.7.8 there are q elements in this subfield. Uniqueness
follows from 1.8.6 (2).
1.8.8. Exercise. (1) Show that the ring R[X]/(X 2 + 1) is isomorphic to the field of
complex numbers.
(2) Show that the ring F2[X]/(X 2 + X + 1) is a field with 4 elements.
(3) Show that the ring F2[X]/(X 3 + X + 1) is a field with 8 elements.
(4) Let K ⊂ L be a field extension of fields of characteristic 0 and let a ∈ L be a root of
an irreducible polynomial f ∈ K[X]. Show that a is a simple root.
(5) Let p be a prime number. Show that Fp is the only ring with p elements.
(6) Let p be a prime number. Show that a ring with p 2 elements is isomorphic to one of
four non isomorphic Z/(p 2 ), Fp × Fp, Fp[X]/(X 2 ), F p 2.