MATH 431 PART 4: INTRODUCTION TO EXTENSION FIELDS
We’ve said a few times that our overarching goal is to find fields where we can completely factor a particular
polynomial, and now we’re ready to build those fields. We will be relying heavily on two major results from the Part
3 notes:
Theorem 1. Let R be a commutative ring with unity. Then R/M is a field if and only if M is a maximal ideal of R.
Theorem 2. Let F be a field and p(x) ∈ F [x]. An ideal hp(x)i =
6 0 of F [x] is maximal if and only if p(x) is irreducible
over F .
Putting these two results together, we know that F [x]/ hp(x)i is a field if and only if p(x) is irreducible . But what
field is it?
1. Extensions of Fields
Let F be a field.
Definition 1. A field E is an extension field of the field F if F is a subfield of E.
Example 1. We’ve already seen a lot of examples of extension fields, but here are two:
(a) Q ⊂ R ⊂ C, so C is an extension field of R, and C and R are both extension fields of Q.
(b) Recall that F (x) is the field of rational functions with coefficients in F , but F is certainly a subfield of F (x)
(thought of as the constant polynomials in F (x)). Similarly, we could construct F (y) as an extension of F , and
F (x, y) (rational functions in two variables) is an extension field of both F (x) and F (y).
We draw subfield diagrams (or extension field diagrams) the same way we drew subgroup diagrams in MATH 430.
Draw the subfield diagrams for these two examples.
Example 2. Consider an element α ∈ E, where E is an extension of F . Then construct the field generated by F and
α, which contains combinations of powers and products of α with elements of F . Thus every element of this field is of
the form
a0 + a1 α + a2 α2 + · · · + an αn + · · ·
where all ai ∈ F . This is, in fact, the smallest field containing both F and α, is denoted F (α), and is an extension of
the field F . Remember: We saw the other day that if α is the root of a polynomial then there is an n ∈ Z+ such
that αn can be expressed in terms of lower powers of α, so F (α) = a0 + a1 α + · · · + an−1 αn−1 | ai ∈ F .
Definition 2. An extension field E of a field F is a simple extension if E = F (α) for some α ∈ E.
2. Roots of polynomials
For most fields F , there are polynomials in F [x] which have no roots in F . On the other hand, in C every polynomial
factors completely as a product of linear polynomials! Is there a middle ground? If I only care about one polynomial,
do I need to go all the way to C to find all of its roots? It turns out, no! As we saw in class previously:
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MATH 431 PART 4: INTRODUCTION TO EXTENSION FIELDS
Theorem 3. Suppose F is a field, E is some extension of F , and p(x) ∈ F [x] is a monic irreducible polynomial
such that α ∈ E is a root of p(x). Then F [x]/ hp(x)i ∼
=
In the last class, we just predicted this theorem. How would we prove this carefully?
Recall the evaluation homomorphism φα : F [x] → E for α ∈ E. Suppose ker φα is nontrivial.
Question: F is a field, so what do we know about every ideal in F [x]? What does this tell us about ker φα ?
Question: We know p(x) is of minimal degree in ker φα . Why can we assume p(x) is monic and irreducible?
Now we consider φα [F [x]]. We know that it is a subring of E because φα is a homomorphism preserving addition,
multiplication, and additive inverses.
Question: What are the elements of φα [F [x]]?
Question: Show that φα [F [x]] is a field. (Show multiplicative inverses! Start by supposing that f (α) ∈ φα [F [x]]
is not zero for f (x) ∈ F [x].)
Question: φα [F [x]] = {f (α) | f ∈ F [x]}. Show that this set is F (α).
Proof (of Theorem 3). Use the Fundamental Homomorphism Theorem with the pieces we constructed above.
MATH 431 PART 4: INTRODUCTION TO EXTENSION FIELDS
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• By Thm 3, Q[x]/ x3 − 7 is isomorphic to what simple extension of Q?
• What is R[x]/ x2 + 1 isomorphic to?
Example 3.
Corollary 4. Suppose that α, β ∈ E are both roots of p(x). Then
Thus we have shown that if a field F has an extension E containing a root of a polynomial p(x) ∈ F [x], then the
field F (α) is such an extension. Our proof assumed that we started with α ∈ E; now we start with a polynomial and
look for a field in which that polynomial has a root. (Remember, by the way, that THIS was our MAIN GOAL in
studying polynomials and fields!)
Theorem 5 (Kronecker’s Theorem). Let F be a field and f (x) a non constant polynomial in F [x]. Then there
exists an extension field E of F and an element α ∈ E such that f (α) = 0.
Proof. Step 1. f (x) has a factorization into irreducibles, so if p(x) is a nonconstant irreducible factor of f (x) then it
is sufficient to find an extension of F containing a root of p(x). (Why?)
Step 2. Since p(x) is irreducible over F , we know that F [x]/ hp(x)i is a field. Let I = hp(x)i. Define ψ : F → F [x]/I
by ψ(a) = a + I for all a ∈ F . ψ is a one-to-one homomorphism so ψ is an isomorphism between F and
ψ[F ] ⊆ F [x]/I. Then F is isomorphic to a subfield of F [x]/I (we identify F with {a + I | a ∈ F }), and we can
conclude that F [x]/I is an extension field of F .
Step 3. The last step is to prove that F [x]/I contains a root of p(x). Consider α = x + I ∈ F [x]/I, and let p(x) =
a0 + a1 x + · · · + an xn . (Now compute p(α) = p(x + I) in F [x]/I.)
Remark 1. Kronecker’s Theorem may seem repetitive after Theorem 3, but in fact they say two really different things.
Kronecker’s Theorem is an existence theorem; it says that for any polynomial over F , there exists a field extension of F
containing any root of that polynomial (and so by iteration we could construct an extension containing all roots of that
polynomial!). The proof is constructive, and says that this field extension is F [x]/ hp(x)i. On the other hand, Theorem
3 gives us an explicit isomorphism so that we have a nicer way to think about these quotient fields F [x]/ hp(x)i where
the roots of p(x) live.
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MATH 431 PART 4: INTRODUCTION TO EXTENSION FIELDS
3. Types of extensions
Definition 3. Let E be an extension field of F and let α ∈ E. We call α algebraic over F if it is the root of some
nonzero polynomial f (x) ∈ F [x]. Otherwise, α is called transcendental over F .
Example 4.
√
2 is algebraic over Q since it is a root of x2 − 2 ∈ Q[x]. π is transcendental over Q, but algebraic over
R (since it is the root of x − π ∈ R[x]).
Example 5. Determine if
p
√
1 + 3 2 is algebraic over Q.
Lemma 6. α ∈ E is algebraic over F if and only if ker φα
and α ∈ E is transcendental over F if and only if ker φα
√
2 is a root of x2 − 2, x3 − 2x, (x2 − 2)(x2 − 1), and many other polynomials. Why do we prefer to use x2 − 2?
Theorem 7. Let E be an extension field of F and let α ∈ E be algebraic over F . Then there exists a unique monic
irreducible nonconstant polynomial p(x) ∈ F [x] of minimal degree such that p(α) = 0.
(“p(x) of minimal degree in F [x]” means that if f (α) = 0 for any other f (x) ∈ F [x], then f (x) divides p(x).)
Definition 4. If p(x) is the unique monic irreducible polynomial of minimal degree with p(α) = 0 guaranteed by
Theorem 7, we call p(x) the minimal polynomial for α over F and we write p(x) = minpol(α). (Our textbook
calls it the irreducible polynomial for α over F and writes p(x) = irr(α, F ).) The degree of α over F is the
degree of its minimal polynomial. (Our textbook denotes this deg(α, F ).)
Remark 2. Notice that we say “minimal polynomial over F ” and “degree over F ” since in each case it matters what
√
√
√
F is. For example, 2 has minimal polynomial x2 − 2 over Q but minpol( 2) = x − 2 over R.
Example 6. Was the polynomial you found in Example 4 the minimal polynomial of
it! What is its degree?
p
1+
√
3
2 over Q? If not, find