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

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 $$M$$.

• Yes this works. Note that the multiplicative structure you put on $K^2$ makes it isomorphic to $K[t]/(t^2-x)$ which is the standard construction. Feb 6, 2019 at 18:29
• Some advice: It would be useful to break up this wall-of-text into paragraphs. Begin and end each chunk of the proof with a high-level description of the claim you're proving, and be clear about whether the chunk is complete or whether it will be justified later. For example, in the sentence "We can verify that $(K^2,+',\cdot')$ is a field...", it's unclear whether you mean "the verification is straightforward and omitted" or "the verification is as follows". The next sentence provides no clue: "Let $(a,b) \neq (0,0)$." Why? Why do we care about $(a,b)$ and its relationship to $(0,0)$? Feb 6, 2019 at 18:46

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 $$L$$.
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 $$K_n$$.