# 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(a2-cb2) belongs to k. Hence (a2 - cb2) 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 (a2-cb2)) is fixed by G and hence belongs to k. Therefore c (a2-cb2) is a square in k.

The result follows.