Unformatted text preview: 330 7. FIELD EXTENSIONS – FIRST LOOK The structure of the splitting ﬁeld can be better understood if we consider the possible intermediate ﬁelds between K and E , and introduce
symmetry into the picture as well.
Proposition 7.4.2. Let K be a ﬁeld, let f .x/ 2 KŒx be irreducible, and
suppose ˛ and ˇ are two roots of f .x/ in some extension ﬁeld L. Then
there is an isomorphism of ﬁelds W K.˛/ ! K.ˇ/ such that .k/ D k
for all k 2 K and .˛/ D ˇ . Proof. According to Proposition 7.3.6, there is an isomorphism
˛ W KŒx=.f .x// ! K.˛/ that takes Œx to ˛ and ﬁxes each element of K . So the desired isomorphism
K.˛/ Š K.ˇ/ is ˇ ı ˛ 1 .
Applying this result to the cubic equation, we obtain the following:
For any two roots ˛i and ˛j of the irreducible cubic polynomial f .x/,
there is an isomorphism K.˛i / Š K.˛j / that ﬁxes each element of K and
takes ˛i to ˛j .
Now, suppose that ı 2 K , so dimK .E/ D 3. Then E D K.˛i / for
each i , so for any two roots ˛i and ˛j of f .x/ there is an automorphism
of E (i.e., an isomorphism of E onto itself) that ﬁxes each element of K
and takes ˛i to ˛j . Let us consider an automorphism of E that ﬁxes
K pointwise and maps ˛1 to ˛2 . What is .˛2 /? The following general
observation shows that .˛2 / is also a root of f .x/. Surely, .˛2 / ¤ ˛2 ,
so .˛2 / 2 f˛1 ; ˛3 g. We are going to show that necessarily .˛2 / D ˛3
and .˛3 / D ˛1 .
Proposition 7.4.3. Suppose K Â L is any ﬁeld extension, f .x/ 2 KŒx,
and ˇ is a root of f .x/ in L. If is an automorphism of L that leaves F
ﬁxed pointwise, then .ˇ/ is also a root of f .x/. Proof. If f .x/ D
0: P fi x i , then P fi .ˇ/i D P
. fi ˇ i / D .0/ D
I The set of all automorphisms of a ﬁeld L, denoted Aut.L/, is a group.
If F Â L is a subﬁeld, an automorphism of L that leaves F ﬁxed pointwise
is called a F –automorphism of L.
The set of F automorphisms of L, denoted AutF .L/, is a subgroup of Aut.L/ (Exercise 7.4.4). ...
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