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I am thinking about my previous question here.

Is there a notion of deleting an element $\alpha$ from a field $F$ that is always well defined? I learned that you cannot always find a proper subfield $E$ such that $E(\alpha)=F$. Take the case of $\mathtt{R}$ and $\alpha=\sqrt{2}$.

However, in the case of $R$, there are many subfields that do not contain the element $\alpha=\sqrt{2}$. Is there a maximal proper subfield? Simply taking unions of subfields would not work as a union of two subfields is not guaranteed to be a field.

I thought if we have two subfields $Q(\beta_1, \dots, \beta_n)$ and $Q(\beta', \dots, \beta_{n'}')$ that do not contain $\sqrt{2}$, we can get a larger subfield $Q(\beta_1, \dots, \beta_n, \beta', \dots, \beta_{n'}')$ that also does not contain $\sqrt{2}$. This was pointed out to be not to be possible in the comments. In some specific cases I think it works, though.

However, my algebra is rusty and I am not sure how to combine subfields that are common extensions of the same underlying field. I know at least one field exists as the algebraic closure of $Q$ exists (take away the imaginary numbers). Certainly a parent subfield exists but in our case will it be proper if they both exclude an element?

My Question: Is there a general notion of field deletion of the element $\alpha$ from the field $F$? I particularly want this notion to undo field extensions, so that would probably be easily gotten.

More specifically, I also want to see the specifics of any such notion in the case of deleting an element of $R$, like $\sqrt{2}$, from $R$.

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  • $\begingroup$ Why does the bigger subfield also not contain $\sqrt{2}$? I don't really see that... Anyway, even if it is true it would strongly depend on the element being $\sqrt{2}$. Take for an other example $\alpha = \sqrt{2} + \sqrt{3}$. Then $\alpha \not \in Q(\sqrt{2})$ and $\alpha \not \in Q(\sqrt{3})$ but we have $\alpha \in Q(\sqrt{2},\sqrt{3})$, if I remember correctly we even have $Q(\sqrt{2},\sqrt{3}) = Q(\alpha)$... $\endgroup$ – Dirk Oct 19 '16 at 12:24
  • $\begingroup$ Ah! Thank you for this correction. $\endgroup$ – abnry Oct 19 '16 at 12:28
  • $\begingroup$ Zorn's lemma ensures such a field always exists, however I think that uniqueness is hopeless. $\endgroup$ – Crostul Oct 19 '16 at 12:46
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Let $K$ be a field an $\alpha \in K$. Let $M = \{L \leq K \mid \alpha \not \in L\}$ be the set of subfields of $K$ that do not contain $\alpha$. Then using Zorn's Lemma we can show that $M$ contains maximal elements (as long as $M$ is nonempty at least), but we won't get a uniqueness of this element, I'm not even sure we might get uniqueness up to isomorphisms or any structure of these maximal elements at all.

The proof using Zorn is quite simple because if we have an ordered sequence of fields $L_1 \leq L_2 \leq L_3 \leq L_4 \leq \ldots$ then also their union is a field.

And an algorithm to construct a counter example for the uniqueness up to isomorphism: Take $K|Q$ a Galois extension of degree $6$ with primitive element $\alpha$. Then $\alpha$ is not contained in any non trivial subfield of $K$. But $K$ has a maximal subfield of index $2$ and one of index $3$ which then both not contain $\alpha$ but are not isomorphic (as they have different dimensions as a $Q$-vectorspace).

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  • $\begingroup$ I am not convinced $M$ is an ordered sequence of fields. Can you give me more? $\endgroup$ – abnry Oct 19 '16 at 12:30
  • $\begingroup$ I am sorry, the set is partially ordered by set inclusion. $\endgroup$ – abnry Oct 19 '16 at 12:31
  • $\begingroup$ Okay, possibly it is unique iff $F=E(\alpha)$? $\endgroup$ – abnry Oct 19 '16 at 12:32
  • $\begingroup$ If I understand this right, by way of K being a galois extension, $K$ is the smallest field (in $R$) that contains $\alpha$. So as you say any proper subfield cannot contain $\alpha$. It is not clear to me why there are maximal subfields of index $2$ and $3$. Can you give me a particular polynomial $K$ is the splitting field of? Is it something like $p(x)=x^6-8$ with primitive $\alpha=\sqrt{2} e^{i \pi/3}$? Sorry for not quite getting it. $\endgroup$ – abnry Oct 19 '16 at 20:21
  • $\begingroup$ It occurs to me though that there might be a fix as it seems in this case the "right" field to choose as the resultant of the deletion of $\sqrt{2}e^{i\pi/3}$ from $K$ would be $Q$ itself. But maybe you'd always get $Q$. $\endgroup$ – abnry Oct 19 '16 at 20:25

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