**189-570A:** Higher Algebra I

## Assignment 5: Corrections.

People mostly did fine one most of the questions.

Many of the solutions to question
6 were longer than they needed to be.

Here is a short solution to question 6.

The equivalence of (i), (ii) and (iii) was explained in class. Assume
that (i), (ii) and (iii) are satisfied, and let G be the Galois
group of L over k. There are two possibilities for G:
(case 1) G is a product of two cyclic groups of order 2, and (case 2)
G is cyclic of order 4.

Let u be a square root of a+bs, and let v be a square root of a-bs.
The elements u,-u,v,-v are roots of an irreducible
polynomial in k[x] of degree 4. Hence G permutes these roots
transitively.

In case 1, G is generated by two elements g and h, g being the permutation
which interchanges u and v, and -u and -v; and
h being the permutation that interchanges u and -u, and v and -v.
Hence the expression uv is fixed by both g and h, and hence by all of G.
Hence uv = sqrt(a^{2}-cb^{2}) belongs to k.
Hence (a^{2} - cb^{2}) is a square in k.

In case 2, G is generated by an element g of order 4 which sends u to v,
v to -u, -u to -v and -v to u.
This element also sends s to -s.
Hence the epression
suv = sqrt(c (a^{2}-cb^{2})) is fixed by G and hence belongs to k.
Therefore
c (a^{2}-cb^{2}) is a square in k.

The result follows.