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 $\alpha \notin F$.

  • $\begingroup$ 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

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 $\mathbb{R}$.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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.

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.