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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 $\alpha \notin F$.

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  • $\begingroup$ Trivial: let $\alpha=1$ and $F=\mathbb{R}$. $\endgroup$ Oct 4 '16 at 20:44
  • $\begingroup$ Whoops, ignore elements of $Q$. $\endgroup$
    – abnry
    Oct 4 '16 at 20:45
  • $\begingroup$ I want $\alpha \notin F$. $\endgroup$
    – abnry
    Oct 4 '16 at 20:48
  • $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Oct 4 '16 at 20:59
  • $\begingroup$ Whoops, no, I am not requiring it to be algebraic. A little rusty with my algebra. $\endgroup$
    – abnry
    Oct 4 '16 at 21:00
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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 $\mathbb{R}$.

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    $\begingroup$ 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. $\endgroup$ Oct 4 '16 at 21:11
  • $\begingroup$ @arctictern A much simpler argument, good point. $\endgroup$ Oct 4 '16 at 21:12
  • $\begingroup$ This is great! Also: It is not clear to me why it has no squadre root. Explain? $\endgroup$
    – abnry
    Oct 4 '16 at 21:25
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    $\begingroup$ @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.) $\endgroup$ Oct 4 '16 at 21:26
  • $\begingroup$ Wonderful. Good argument $\endgroup$
    – abnry
    Oct 4 '16 at 21:27
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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.

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

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