I need a hint to solve exercise 13.2.9 in Dummit and Foote. Suppose $F$ is a field of char not equal to 2. Suppose $a^2 -b$ is a square where $a,b \in F$ and $b$ is not a square.
Show $\sqrt{a + \sqrt{b}} =\sqrt{m} +\sqrt{n}$ for some $m,n \in F$.
I've reduced to $(a+\sqrt{b})(a-\sqrt{b})=(\sqrt{m}+\sqrt{n})^{2}$ but dont know how to proceed
Best Answer
Hypothetically, if $\sqrt{a + \sqrt{b}} = \sqrt{m} +\sqrt{n}$ then $a + \sqrt{b} = m + n + 2 \sqrt{m n}$ so it could be that $a = m + n$ and $b = 4 m n$, then we would further have $a^2 - b = (m - n)^2$.
So let's say given $a,b$ we define $m = \tfrac{1}{2}(a + \sqrt{a^2 - b})$, $n = \tfrac{1}{2}(a - \sqrt{a^2 - b})$. These are certainly elements of the field due to the condition of $a^2 - b$ being a square, furthermore multiplying it out proves that it works.