Fields closed up to degree $n$

Question

Are there fields that are not algebraically closed fields but such that any quadratic polynomial has a root?

Suppose now $n$ is a fixed natural number, $n\geq 2$. Is there a field $K_n$ such that every polinomial of degree $\leq n$ with coefficients in $K_n$ has a root in $K_n$, but $K_n$ is not algebraically closed?

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I was thinking that perhaps a tower of algebraic extensions of $\mathbb{Q}$ might work, but then I realized that in order to be "quadratically closed" I might have to close under all extensions of degree $2^k$ for some $k$.

Is $K_2=\bigcup \left\{\mathbb{Q}[\alpha]: [\mathbb{Q}(\alpha):\mathbb{Q}]=2^k\text{ for some $k$}\right\}$ going to work as a field that is quadratically closed but not algebraically closed?

Answer 1

The probably must famous example are the constructable numbers, i.e. the lengths that can be constructed using a straight-edge, a compass, and an already drawn line segment of length $1$. You can add, subtract, multiply, divide and solve any quadratic equation with constructions, as long as the coefficients are constructible, but you cannot take cube roots.

For the general situation, let $n\in \Bbb N$, and let $K = K_0$ be a field not of characteristic $n$. Then, let $K_1 \subseteq \overline K$ be the smallest subfield containing $K$ and the roots of all polynomials of degree $n$ with coefficients in $K$. Next, define $K_2\subseteq \overline K$ be the smallest subfield containing $K_1$ and the roots of all degree-$n$ polynomials with coefficients in $K_1$.

Keep going like that, and you will get a chain $K= K_0\subseteq K_1\subseteq\cdots$ of fields. The union $K_\omega$ of all these fields is still a field, and it will contain the zeroes of any $n$-th degree polynomial with coefficients in $K_\omega$. The proof of these two statements both revolve around the same fact: For any given addition, subtraction, multiplication, division and polynomial, there are only finitely many numbers involved, so the problem is entirely contained in some $K_i$ for a finite $i \in \Bbb N$.

Note that $K_\omega$ may be algebraically closed, even though $K$ isn't. For a simple example, set $K = \Bbb R$ and $n = 2$. In that case $K_1 = K_2 = \cdots = \Bbb C$. However, in the case of, for instance, $K = \Bbb Q$, this process cannot end in $\overline {\Bbb Q}$, since that would mean that $\sqrt[n+1]2 \in K_\omega$, which again means that there is an $i\in \Bbb N$ such that $\sqrt[n+1]2 \in K_i$. How could that have gotten there?

The answer is, of course, that it couldn't. Over $\overline{\Bbb Q}$, all $n$-th degree factors of $x^{n+1} - 2$ have $\sqrt[n+1]2$ (and powers thereof, with some complex $(n+1)$-th roots of $1$ and $-1$ thrown into the mix) as coefficients, which means that there cannot ever be a smallest $i$ for which $\sqrt[n+1]2\in K_i$.

Answer 2

Yes, such a quadratically closed field field exists.
Start with an arbitrary field $K=K_0$ of characteristic $\neq 2$ and adjoin all square roots of elements of $K$.
In other words consider the splitting field of the family ($X^2-q)_{q\in K}$ and obtain the field $K_1=K(\sqrt q\vert q\in K)=K(\sqrt K)$.
Define inductively $K_{n+1}=K_n(\sqrt K_n)$ and obtain an increasing sequence of fields $$K=K_0\subset\cdots \subset K_n\subset\cdots$$ The union $K_\text{quad}=\cup_{n=0}^\infty K_n$ is a field (like all increasing union of fields) closed under extraction of square roots but $K_\text{quad}$ is not algebraically closed in general: indeed any element of $K_\text{quad}$ has degree $2^n$ for some $n$ .