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Let $K / F$ and $L / F$ be two extension of fields, where $K / F$ is furthermore monogenic, such that $K$ and $L$ are isomorphic (as fields). Can I state that $L / F$ is monogenic too? If not, can I add extra conditions to $K$, $L$ and $F$ to get it? My attempt has no result, but I suppose that $\theta : K \to L$ is an isomorphism and $K = F(x)$, one candidate for $L$ seems to be $F(\theta(x))$. So if I take an arbitrary element $y$ in $L$ and I suppose that $y \notin F$, how can I see that $y \in F(\sigma(x))$? Is it possible? Thank you very much in advance.

joseabp91
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1 Answers1

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No, this is extremely false. For instance, if $k$ is any field and $F=k(X_1,X_2,\dots)$ is a field of rational functions in infinitely many variables, then the fields of rational functions $K=F(Y)$ and $L=F(Y,Z)$ are isomorphic but $K$ is monogenic over $F$ and $L$ is not.

It is true if you assume $K$ and $L$ are isomorphic not just as fields but as fields over $F$. That is, assume there is an isomorphism $\theta:K\to L$ which is the identity on $F$ (or actually, it would suffice to have $\theta(F)\supseteq F$). Then the proof is trivial: if $K=F(a)$, then $L=\theta(F(a))=\theta(F)(\theta(a))=F(\theta(a))$.

Eric Wofsey
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  • Is my statement false even if the extensions $K / F$ and $L / F$ are finite? – joseabp91 Aug 05 '18 at 19:01
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    Yes; if $k$ has characteristic $p>0$, you can take $F$ as in my example and $K=F(x_1^{1/p})$ and $L=F(x_1^{1/p},x_2^{1/p})$. – Eric Wofsey Aug 05 '18 at 19:03
  • In your example, are $x_1 , \ldots$ the functions in infinitely variables? – joseabp91 Aug 05 '18 at 19:12
  • I don't know what you mean by "the functions". $x_1,x_2,\dots$ are just an infinite collection of variables, and $F$ is the field of formal rational functions in this variables with coefficients in $k$. – Eric Wofsey Aug 05 '18 at 19:14
  • What do you mean with "field of formal rational functions"? – joseabp91 Aug 05 '18 at 19:18
  • Just the field of fractions of a polynomial ring. – Eric Wofsey Aug 05 '18 at 19:19
  • But what are $y$ and $z$? I can take $z$ in $F$ too. – joseabp91 Aug 05 '18 at 19:35
  • They are again formal variables. Normally, when someone uses a variable like $x$ or $y$ or $t$ in talking about a field extension like $F(y)$, this refers to a formal variable, unless some other meaning is clear from context. – Eric Wofsey Aug 05 '18 at 19:35
  • But when I wrote $K = F(x)$, I meant that $x \in K$. So in this case $x$ is not a formal variables. – joseabp91 Aug 05 '18 at 19:37
  • For example the extension $\mathbb{Q}(\sqrt{2})$. – joseabp91 Aug 05 '18 at 19:40
  • What's your point? You used a variable in one way, and I used some different variables a different way. – Eric Wofsey Aug 05 '18 at 19:40
  • My $x$, when I wrote $K = F(x)$, is not a variable. It is not the field of fractions. It is simply the minimal field which contains as $F$ as $x \in K$. – joseabp91 Aug 05 '18 at 19:46
  • Yes, I understand that and always have. I continue to not know what your point is. – Eric Wofsey Aug 05 '18 at 19:47
  • What do you mean exactly? It is a fixed point in $K$. – joseabp91 Aug 05 '18 at 19:48
  • Oh, I think there is a language barrier here. When I refer to "your point" I mean "the purpose of what you are saying". See https://www.phrasemix.com/phrases/whats-the-point-of-something. – Eric Wofsey Aug 05 '18 at 19:51
  • I am sorry. I think that I have just got what you meant. My propose is see that my statement is indeed false. I have a question. What is a formal variable for you? – joseabp91 Aug 05 '18 at 19:57
  • An indeterminate, an element which is transcendental over the base. When I say $K=F(y)$, all I mean is that $K$ is the field of fractions of a polynomial ring in one variable over $F$. – Eric Wofsey Aug 05 '18 at 20:01
  • In your case, $y$ is not an element in $K$, right? – joseabp91 Aug 05 '18 at 20:02
  • It is an element of $K$! $K$ is the field of fractions of the ring of polynomials in the variable $y$ with coefficients in $F$. One such polynomial is $y$ itself. – Eric Wofsey Aug 05 '18 at 20:04
  • Can I take $K = \mathbb{Q}(\sqrt{2})$ and $L = \mathbb{Q}(\sqrt{2} , \sqrt[3]{5})$? – joseabp91 Aug 05 '18 at 20:56
  • No. Any finite field extension over a field of characteristic $0$ (or more generally any finite separable extension) is monogenic, so you will never get any example that way. (And in any case, those fields are not isomorphic...) – Eric Wofsey Aug 05 '18 at 20:57
  • Can you give me an specific example, please? – joseabp91 Aug 05 '18 at 20:58
  • Example of what? – Eric Wofsey Aug 05 '18 at 20:58
  • A counterexample which shows that my statement is false, taking a specific fields and points. – joseabp91 Aug 05 '18 at 20:59
  • I have given a specific example. The only thing unspecific is that I have not said what $k$ is, but you can plug in any field at all for $k$. – Eric Wofsey Aug 05 '18 at 20:59
  • Can I take $k = \mathbb{Q}$? – joseabp91 Aug 05 '18 at 21:01
  • Sure. ${}{}{}{}{}$ – Eric Wofsey Aug 05 '18 at 21:05
  • Can you give me a specific $F$, please? – joseabp91 Aug 05 '18 at 21:07
  • $F$ is the field of fractions of the ring of polynomials in the variables $x_1,x_2,\dots$ with coefficients in $\mathbb{Q}$. – Eric Wofsey Aug 05 '18 at 21:09
  • Can you give me a specific variables $x_1 , x_2 , \ldots$? – joseabp91 Aug 05 '18 at 21:11
  • Do you know what a polynomial ring is? For a very explicit definition, see https://math.stackexchange.com/questions/1838860/polynomial-ring-with-arbitrarily-many-variables-in-zf – Eric Wofsey Aug 05 '18 at 21:13
  • For a concrete field which is isomorphic to this $F$, you can let $a_1,a_2,\dots\in\mathbb{C}$ be algebraically independent over $\mathbb{Q}$ and let $F$ be the subfield of $\mathbb{C}$ generated by $a_1,a_2,\dots$. – Eric Wofsey Aug 05 '18 at 21:14
  • I know what it is. I usually denote by $X$ if I want to denote it as a variable, but $x$ as an element. Then are $x_1 , x_2 , \ldots$ elements in $K$ or variables? – joseabp91 Aug 05 '18 at 21:15
  • They both variables and elements of $F$ (and thus also of $K$ since $K$ is an extension of $F$). If you usually use capital letters for variables, then $F$ is what you would write as $\mathbb{Q}(X_1,X_2,\dots)$. – Eric Wofsey Aug 05 '18 at 21:17
  • No, they are not the same. Usually when you write $F[X]$, it is the ring of polynomials with coefficients in $F$, so $X$ is here a variables and not an element. But if you take $K = F(x)$, then $x$ is an element in $K$, it is the minimal field which contains as $F$ as $x$ and it coincides with the set of fractional polynomials evaluated in $x$. – joseabp91 Aug 05 '18 at 21:19
  • I am aware that this is a convention that is sometimes used. It is not universal, and I was not using it. Just turn all my variables into capital letters, if that makes you happier. – Eric Wofsey Aug 05 '18 at 21:20
  • Note that even when one does write $F[X]$, though, $X$ is an element of $F[X]$ (and also of its fraction field $F(X)$). Being an element and being a variable are not mutually exclusive. – Eric Wofsey Aug 05 '18 at 21:21
  • I have just understood the example that you want to share. I have not contemplated these fields before. How can I show that $\mathbb{Q}(X_1 , X_2 , \ldots)$ is a field? – joseabp91 Aug 05 '18 at 21:57
  • Well, it is usually defined as the field of fractions of $\mathbb{Q}[X_1,X_2,\dots]$. – Eric Wofsey Aug 05 '18 at 22:00
  • How can I show that $\mathbb{Q}[X_1 , X_2 , \ldots]$ is an integral domain? – joseabp91 Aug 05 '18 at 22:03
  • There are lots of ways. Here's on: any element contains only finitely many of the variables, so any two elements can be multiplied in a finite-variable polynomial ring $\mathbb{Q}[X_1,\dots,X_n]$. So by induction, it suffices to show that if $R$ is an integral domain then so is $R[X]$. You can prove that by looking at what happens to the highest degree terms when you multiply two polynomials. – Eric Wofsey Aug 05 '18 at 22:56