Every finite field has an extension in which every element has a square root

Question

I know every field has an algebraic closure, I just wanted to prove this statement separately using "simpler" methods.

Let $(K,+,\cdot)$ be a finite field. First suppose there exists an $x \in K$ such that for every $y \in K$ we have $y^2 \neq x$. Define the operations $+'$ and $\cdot'$ on $K^2$ with $(a,b)+'(c,d) = (a+c,b+d)$ and $(a,b)\cdot'(c,d) = (ac+bdx,ad+bc)$. We can verify that $(K^2,+',\cdot')$ is a field, with $(1,0)$ being the multiplicative neutral element etc. Let $(a,b) \neq (0,0)$. Suppose the vectors $(a,b)$ and $(bx,a)$ are linearly dependent, i.e. there exists a $c \in K$ such that $cb = a$ and $bx = ca = c^2b$. If $b = 0$, then so is $a$, contradicting $(a,b) \neq (0,0)$. If $b \neq 0$, then we obtain $x = c^2$, contradicting our assumption $\forall y \in K: y^2 \neq x$. Thus, the vectors $(a,b)$ and $(bx,a)$ are linearly independent which means that the linear equation system $(a,b)(c,d) = (ac+bdx,ad+bc) = (1,0)$ for $(a,b) \neq (0,0)$ has a unique solution, proving that $(K^2,+',\cdot')$ is a field. Obviously, $\{(a,0)\;|\;a \in K\}$ is a subfield of $K^2$ that is isomorphic to $K$ so that we may replace $(a,0)$ with $a$. We also see that $(0,1)(0,1) = (x,0) = x$ which means that $x$ has a square root in $K^2$. Since $K$ is finite, there are only finitely many elements that may not have a square root in $K$, so we can repeat constructing field extensions until we obtain a field extension $L \geq K$ where every element in $K$ has a square root in $L$. $L$ is still a finite field. Now, let $K_0 := K, K_1 := L$, let $K_2$ be the field extension of $K_1$ where every element of $K_1$ has a square root in $K_2$ and so on. Let $M := \cup_{n\in\mathbb{N}} K_n$. Since $K_0 \subset K_1 \subset K_2 \subset ...$, for every $a,b \in M$, there exists a smallest $n \in \mathbb{N}$ such that $a,b \in K_n$. For these $a,b$, define operations $+$ and $\cdot$ on $M$ as the respective $+$ and $\cdot$ in $K_n$. It is easy to show that $(M,+,\cdot)$ is a field. Let $x \in M$. Then there exists an $n \in \mathbb{N}$ such that $x \in K_n$. By construction of $K_{n+1}$, x has a square root y in $K_{n+1} \subset M$. Thus, every $x \in M$ has a square root in $M$.

Answer 1

Yes, that works. Some additional notes:

First suppose there exists an $x\in K$ such that for every $y\in K$ we have $y^2\neq x$.

Every finite field that doesn't have characteristic $2$ has non-square elements. Every finite field of characteristic $2$ has all of its elements squares, and $x\to x^2$ is an automorphism of such a field.

... so we can repeat constructing field extensions until we obtain a field extension $L\ge K$ where every element in $K$ has a square root in $L$.

Once will suffice. For finite fields, there is exactly one extension (up to isomorphism) of each degree - so when we adjoin one square root, we get all of the others for free.

This whole construction, taking a chain of extensions and their union, is a case of the direct limit - the same construction we would use to construct the algebraic closure of the finite field we started with. The difference is that we used only extensions of power-of-2 order instead of extensions of all finite orders, which means we get a total order instead of a partial order on our $K_n$.