
68 Field Extensions
Q(i,−i,
√
5,−
√
5). As written, it appears to require the adjunction of four new elements.
Clearly just two, i and
√
5, suffice. But we claim that in fact only one element is
needed, because L = L
0
where L
0
= Q(i +
√
5), which is obviously simple. To prove
this, it is enough to show that i ∈ L
0
and
√
5 ∈ L
0
, because these imply that L ⊆ L
0
and L
0
⊆ L, so L = L
0
. Now L
0
contains
(i +
√
5)
2
= −1 + 2i
√
5 + 5 = 4 + 2i
√
5
Thus it also contains
(i +
√
5)(4 + 2i
√
5) = 14i −2
√
5
Therefore it contains
14i −2
√
5 + 2(i +
√
5) = 16i
so it contains i. But then it also contains (i +
√
5) −i =
√
5. Therefore L = L
0
as
claimed, and the extension Q(i,−i,
√
5,−
√
5) : Q is in fact simple. ...