# A question about algebraically closed fields

A field $\mathbb{K}$ is said to be algebraically closed in practice if every polynomial over $\mathbb{K}$ of positive degree less than or equal to $10^{10}$ has zero belonging $\mathbb{K}$. The question arises: is it possible that an algebraically closed in practice field is not algebraically closed?

PS. The question still remains open in characteristic 0.

• Easily. Take the splitting field over a finite field ($\mathbb F_p$) of the polynomial $f(x)$ obtained by multiplying all polynomials of degree $\le 10^{10}$ together. – Dustan Levenstein Jul 25 '14 at 14:38
• Why is the splitting field algebraically closed in practice? – user64494 Jul 25 '14 at 14:42
• @ Dustan Levenstein: Can you explain your suggestion in detail? – user64494 Jul 25 '14 at 14:44
• Who has suggested this definition of an algebraically closed field in practice? An applied mathematician? Why $10^{10}$? – Martin Brandenburg Jul 25 '14 at 14:50
• If you take the union of all $\mathbb F_{p^{(N!)^i}}$, for $i>0$, then you get an algebraically closed in practice field which is not algebraically closed. It is algebraically closed in practice because every polynomial with coefficients in the field has coefficients in some fixed $\mathbb F_{p^{(N!)^i}}$, which then splits in $\mathbb F_{p^{(N!)^{i+1}}}$. It is not algebraically closed because it does not contain $\mathbb F_{p^q}$ for $q$ a prime bigger than $N$. – Dustan Levenstein Jul 25 '14 at 14:59

Let me expand on Dustan Levenstein's answer in the comments: inside the algebraic closure $\overline{\mathbb{F}_p}$ of $\mathbb{F}_p$, consider the union $K = \bigcup_i \mathbb{F}_{p^{(N!)^i}}$, for $N = 10^{10}$. Since this is a nested union, $K$ is a field. If $f(x) \in K[x]$ is a polynomial of degree $\le 10^{10}$, then $f(x) \in \mathbb{F}_{p^{(N!)^i}}[x]$ for some $i$ (since $f$ has finitely many coefficients). Setting $q = p^{(N!)^i}$, we have that $f(x) \in \mathbb{F}_q[x]$ has an irreducible factor of degree $\le 10^{10}$, and thus splits in $\mathbb{F}_{q^(N!)} = \mathbb{F}_{p^{(N!)^{i+1}}} \subseteq K$.

However, $K$ is not algebraically closed - if $q > N!$ is prime, then $\mathbb{F}_{p^q} \not \subseteq K$. If it were, then $\mathbb{F}_{p^q} \subseteq \mathbb{F}_{p^{(N!)^i}}$ for some $i$ (since $\mathbb{F}_{p^q}$ is a finite field), but $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^e}$ iff $d \mid e$.

• You wrote "for $N=10^{10}!$ the splitting field of $x^{p^N} - x$ over $F_p$ is algebraically closed in practice". Unfortunately, an algebraically closed in practice field cannot be finite. – user64494 Jul 25 '14 at 17:58
• @ zcn :Let us consider the polynomial $f(x):=x^2-x$ over a finite $\mathbb{K}$. It glues together $0$ and $1$. Therefore, there exists $c \in \mathbb{K}$ s. t. $c \notin f(\mathbb{K})$. – user64494 Jul 25 '14 at 18:21
• @ zcn : pay attention to $f(x)-c$ . – user64494 Jul 25 '14 at 18:31
• @ zcn : Can you explain why $f(x)$ has an irreducible factor of degree less than or equal to $10^{10}$? Why your K is not algebraically closed? – user64494 Jul 25 '14 at 19:28
• @user64494: For your first question: $K[x]$ is a PID, hence a UFD, for any field $K$. For the second: if $K$ were algebraically closed, then it would have to equal $\overline{\mathbb{F}_p}$, which it does not – zcn Jul 25 '14 at 20:27

Here's a construction that works in characteristic $0$. I'm really just expanding on rschwieb's suggested strategy, so I hope it qualifies as a serious answer.

Given $N$ (in this case $10^{10}$), call a subfield $K\subseteq\mathbb{C}$ a "small" field if there is a chain of fields $$\mathbb{Q}=K_0\subseteq K_1\subseteq\dots\subseteq K_n=K$$ where the degree of the extension $K_{i+1}/K_i$ is less than or equal to $N$ for all $i$.

If $$\mathbb{Q}=L_0\subseteq L_1\subseteq\dots\subseteq L_m=L$$ is another such chain, where $L_{i+1}=L_i(\gamma_i)$ is a simple extension (which we can assume, either using the fact that every finite extension of fields of characteristic $0$ is simple, or in a more elementary way by refining the chain of fields to adjoin one element at a time) then so is $$\mathbb{Q}=K_0\subseteq K_1\subseteq\dots\subseteq K_n\subseteq\langle K, L_1\rangle\subseteq\dots\subseteq\langle K,L\rangle,$$ since $$[\langle K,L_{i+1}\rangle:\langle K,L_i\rangle]=[\langle K,L_i\rangle(\gamma_i):\langle K,L_i\rangle]\leq [L_i(\gamma_i):L_i]=[L_{i+1}:L_i]\leq N,$$ (because the minimal polynomial of $\gamma_i$ over $\langle K,L_i\rangle$ divides its minimal polynomial over $L_i$), and by induction the join of finitely many small fields is small, and the union of all small fields is a field $\mathbb{K}$.

If $\alpha\in\mathbb{C}$ is a root of a polynomial $p(t)$ of degree less than or equal to $N$ over $\mathbb{K}$, then all of the coefficients of $p(t)$ are contained in some small field $K$, and the degree of $K(\alpha)$ over $K$ is less than or equal to $N$, so $K(\alpha)$ is small, and so $\alpha\in\mathbb{K}$. Therefore $\mathbb{K}$ is "algebraically closed in practice".

Let $p$ be a prime greater than $N$, and let $\beta\in\mathbb{C}$ be an algebraic number whose minimal polynomial over $\mathbb{Q}$ has degree $p$. Then $\beta$ is contained in no small field, and so $\mathbb{K}$ is not algebraically closed.

• @ Jeremy Ricard: Could you kindly explain "the union of all small fields is a field $\mathbb{K}$" in detail? – user64494 Jul 28 '14 at 15:32
• @user64494 If $x\in K$ and $y\in L$, where $K$ and $L$ are small, then $x+y,xy\in\langle K,L\rangle$, which I showed was a small field. So $\mathbb{K}$ is closed under addition and multiplication. As a union of fields, it's also closed under taking additive or multiplicative inverses. – Jeremy Rickard Jul 28 '14 at 15:38
• @ Jeremy Rickard: What do you mean by $\langle K,L \rangle$? Why is the set of all the "small" fields finite? – user64494 Jul 28 '14 at 16:08
• @user64494 $\langle K,L\rangle$ is the smallest subfield of $\mathbb{C}$ containing both $K$ and $L$. The set of all "small" fields isn't finite (and so $\mathbb{K}$ is not a finite extension of $\mathbb{Q}$, and not a "small" field by my definition). But all that matters is that any finite set of elements of $\mathbb{K}$ is contained in some "small" field. – Jeremy Rickard Jul 28 '14 at 16:21
• @ Jeremy Rickard: Why does $\langle K,L\rangle$ exist? Why is it "small" if $K$ and $L$ are "small"? – user64494 Jul 28 '14 at 16:35

Phenomena like this can occur.

To see this, we can take a baby step and try for $2$ before we try $10^{10}$. There are fields called quadratically closed fields for which every element has a square root. Using that, you can rewrite any monic polynomial of degree $2$ in the form $(x+a)^2-b$ by completing the square. Then it factors into $((x+a)-b')((x+a)+b')$, after which you have a root in the field. So for this type of field, all polynomials of degree no greater than $2$ have a root.

As suggested in the comments, it's probably possible that you can take a field, look at the tower of extensions over the field which split polynomials of low degrees, and argue that the union is algebraically closed in practice.

• The question is for $10^{10}$, not for $2$. I will be waiting for a serious answer. Thank you anyway. – user64494 Jul 25 '14 at 15:30
• Wow, you might want to check your tone there. – Dustan Levenstein Jul 25 '14 at 15:33
• Dear @user64494 : When you said "You may replace $10^{10}$ by $10$" in your comments above, I thought you were open to such a demonstration. I guess you weren't being serious. Maybe in your next pass you can read the rest of the post which suggests a strategy for whatever $n$ you pick. – rschwieb Jul 25 '14 at 16:03