# Does there exist $\alpha \in \mathbb{R}$ and a field $F \subset \mathbb{R}$ such that $F(\alpha)=\mathbb{R}$?

I was thinking what would be the opposite of a field extension and I suppose it might be this: A field $F$ can have the element $\alpha$ deleted if there exists a subfield $E=F$ such that $E(\alpha)=F$. You might call $F\setminus(\alpha)$ something like a deletion. The thing about this is that you can then undo any field extension. That is, $F(\alpha) \setminus (\alpha)=F$. Does this concept have a name?

So naturally, I am wondering if any elements can be "deleted" in this manner from $\mathbb{R}$. I realized that for $\alpha=\sqrt{2}$, there is no such field $E$ such that $E(\alpha)=\mathbb{R}$. If there was, there would be $a,b \in E$ such that $2^{1/4}=a+b\sqrt{2}$ as $\mathbb{R}$ is an extension of $E$ and the fourth root is in $\mathbb{R}$. Squaring, and with some algebra (being careful with avoiding division by zero), one gets the contradiction that $\sqrt{2} \in E$.

I haven't thought this through carefully, but I believe for any algebraic number we have a similar failure. That is, there is a polynomial of degree $n$, $p \in E(x)$, with $p(\alpha)=0$. Then in particular $\alpha^{1/k} = p_k(\alpha)$ for $p_k$ of degree $n$. Raising to the $k$th power we get $\alpha = q_k(\alpha)$ for $q_k$ polynomial of degree $n$. This gives a linear system in $\alpha^k$ and my hunch it is nonsingular or the singular cases can be dealt with.

The above is for algebraic numbers. I think the above can be used to show, for instance, that $\pi$ cannot work as it is likely algebraic over $E$ as $E$ contains many transcendental numbers. This is much more murky to me but makes me think that there is no element you can delete from $\mathbb{R}$.

So my question: Does there exist $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ and a field $F \subset \mathbb{R}$ such that $F(\alpha)=\mathbb{R}$? More specifically, I want $F$ with $\alpha \notin F$.

• Trivial: let $\alpha=1$ and $F=\mathbb{R}$. Oct 4 '16 at 20:44
• Whoops, ignore elements of $Q$. Oct 4 '16 at 20:45
• I want $\alpha \notin F$. Oct 4 '16 at 20:48
• Are you requiring $\alpha$ to be algebraic over $F$? Your use of $F[\alpha]$ rather than $F(\alpha)$ suggests that you do, but I suspect it isn't what you meant. Oct 4 '16 at 20:59
• Whoops, no, I am not requiring it to be algebraic. A little rusty with my algebra. Oct 4 '16 at 21:00

No such subfield $F$ exists.

First, $\mathbb{R}$ cannot be of finite degree over $F$, else $\mathbb{C}$ would be of finite degree over $F$, and this is only possible if $F=\mathbb{R}$ or $\mathbb{C}$ by the Artin-Schreier theorem.

Thus, if such a field existed, then $\mathbb{R}$ would be of infinite degree over $F$, and $a$ would be transcendental over $F$. But then the field $F(a)$ would have many non-trivial automorphisms (for instance, the ones fixing all elements in $F$ and sending $a$ to any degree $1$ polynomial in $a$); since we know that the only field automorphism of $\mathbb{R}$ is the identity, then this is not possible.

A simpler argument pointed out by arctic tern is that neither of $a$ and $-a$ have a square root in $\mathbb{R}$, which is impossible for elements of $\mathbb{R}$.

• Also, in $F(a)$ the element $a$ has no square root, but neither does $-a$ since $F(a)=F(-a)$. Unlike in $\mathbb{R}$, where for all $a$, one of $a$ or $-a$ has a square root. Oct 4 '16 at 21:11
• @arctictern A much simpler argument, good point. Oct 4 '16 at 21:12
• This is great! Also: It is not clear to me why it has no squadre root. Explain? Oct 4 '16 at 21:25
• @abnry Suppose $p(a)/q(a)\in F(a)$ squared to $a$, so then $p(a)^2=aq(a)^2$, but the degrees on the left and right side of that equality are opposite parity (even and odd). (Or you can do the "so $a$ is a divisor of $p(a)$ and $q(a)$ contradicting $p/q$ in lowest terms" trick used to prove $\sqrt{2}$ is irrational.) Oct 4 '16 at 21:26
• Wonderful. Good argument Oct 4 '16 at 21:27

In fact, $\mathbb R$ does not have any subfield of finite index - this is a direct consequence of the Artin-Schreier theorem, which (in addition to other things) states that if $C$ is an algebraically closed field and $F$ is a proper subfield such that the degree $[C : F]$ is finite, then $C = F(\sqrt{-1})$. Assuming that $\mathbb R$ had a proper subfield $E$ such that $\mathbb R = E(\alpha)$ for an $\alpha$ algebraic over $E$ implies that the degree $[\mathbb R : E]$, and thus $[\mathbb C : E]$, is finite. Since $\mathbb C$ is algebraically closed, Artin-Schreier gives the result that $\mathbb C = E(\sqrt{-1})$, so that $E \subset \mathbb R \subset \mathbb C$ with $[\mathbb C : E] = 2$. This implies $E = \mathbb R$, which is a contradiction.

See this writeup by Keith Conrad for more details, and a proof of the Artin-Schreier theorem.

The case when $\alpha$ is not algebraic over $E$ has been covered by Pierre-Guy Plamondon in his answer - I will not repeat his argument here.

• Thanks. For me to accept this answer I need more details on the construction in the transcendental case. Oct 4 '16 at 21:02
• My bad - no such field exists, and the construction doesn't work. I was having hallucinations. Oct 4 '16 at 21:07
• No problem. I'm very rusty on my algebra so you could have fooled me! Oct 4 '16 at 21:23