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I am trying to generalize a problem that I came across previously.

$\mathbf{ Problem:}$ Are the fields $\mathbb{Q}$ and $\mathbb{Q[\sqrt2]}$ isomorphic?

$\mathbf{Generalisation:}$ Let $F$ and $K$ be two fields of characteristic $0$ and such that $F \subset K$. If there exists a polynomial (of finite degree) $p(x) \in F[x]$ with coefficients from copies of $\mathbb{Q}$ such that $F$ does not contain any root of $p(x)$ but $K$ contains every root of it, then $F\ncong K$.

Proof Attempt:

Any field $F$ with $\mathrm{char \ F=0 }$ contains a copy of $\mathbb{Q}$. So, $\mathbb{Q}$ is isomorphic to a subfield of $F$ (thereby $K$).

Now, let us assume that despite of the presence of such a polynomial $p(x)$, $F \cong K$. Let $p(x)=a_0x^n+a_{1}x^{n-1}+...+a_{n-1}x+a_n$. We have $p(\beta)=0$, for some $\beta \in K$ but $\beta \notin F$.

Let $\phi: F \mapsto K$ be such an isomorphism. For some $x \in F$,

$\beta=\phi(x) \implies a_r\beta^{n-r}= \phi(a_rx^{n-r})$ $ \ \ \ \ \ ....(*) $ . [This holds due to the fact that any such isomorphism will fix $\mathbb{Q}$, i.e. $\phi(m)=m$ , $\forall m\in \mathbb{Q}$].

By $(*)$, running $r\in \{1,2,...,n\}$ and adding them gives,

$p(\beta)=0=\phi(p(x)) \implies p(x)=0$. By our assumption, $x \in F$, but $p(x)\neq 0 \ \forall \ x\in F$. A contradiction.

Is the formulation of the general statement correct? Is my "Proof" correct? Kindly verify.

Subhasis Biswas
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    Since $\Bbb Q[x] \subseteq F[x] \subseteq K[x]$, then $p(x)$ must have exactly the same number of roots in $F$ as it has in $K$ or the fields can't be isomorphic. So if $F$ contains no roots of $p(x)$, your conclusion follows if $K$ contains even one root of $p(x)$. – Robert Shore Sep 11 '19 at 07:54
  • @RobertShore Now, from that polynomial whose at least one root is missing in $F$, can't we find another polynomial whose all roots are missing in $F$ but $K$ contains all of its roots? – Subhasis Biswas Sep 11 '19 at 07:57
  • I'm not sure. I'm just pointing out that you are using a stronger assumption than you need to reach your conclusion. Let $F=\Bbb Q, K= \Bbb Q[\sqrt[3]{3}], p(x)=x^3-3$. – Robert Shore Sep 11 '19 at 08:00
  • @RobertShore I agree. Initially I intended to prove the even more general case you mentioned, but then I realized that I am yet to be familiar with the tools required to prove it. Or maybe I am missing something obvious. I will try to prove it later on. – Subhasis Biswas Sep 11 '19 at 08:13

2 Answers2

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Looks good. You could also just take a nonlinear irreducible factor of $p(X), p'(X)$ that $\beta$ is a root of that so $K$ is a nontrivial field extension of $F.$ Any we know from linear algebra or what not that an isomorphism can't exist.

green frog
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You don't need the full strength of this assumption. All you need is one root in $K$.

Assume $\exists p(x) \in \Bbb Q[x]$ such that $p(x)$ has at least one root $r \in K$ but $p(x)$ has no roots in $F$. Then assume $f:F \to K$ is an isomorphism. Since $f(1)=1$, it follows that $\forall a \in \Bbb Q, f(a)=a,$ so $f(p(x))=p(x)$. But $p(r)=0 \Rightarrow 0=f^{-1}(0)=f^{-1}(p(r))=p(f^{-1}(r)) \Rightarrow f^{-1}(r)$ is a root of $p$ in $F$, contradicting our hypothesis.

Robert Shore
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  • Won't it be $f(p(x))=p(f(x))$? – Subhasis Biswas Sep 11 '19 at 16:42
  • You could use that as an intermediate step, but the point is that $f$ is necessarily the identity on $\Bbb Q$. – Robert Shore Sep 11 '19 at 16:45
  • Correct if wrong: Since the indeterminate $x$ is from $\mathbb{Q}$, $f(p(x))=p(x)$. Right? – Subhasis Biswas Sep 11 '19 at 17:01
  • No, $x$ is an indeterminate. What's correct is that since the coefficients of $p(x)$ are from $\Bbb Q, \forall x \in F~ f(p(x))=p(f(x)).$ – Robert Shore Sep 11 '19 at 20:39
  • But in your answer: "Then assume $f:F \to K$ is an isomorphism. Since $f(1)=1$, it follows that $\forall a \in \Bbb Q, f(a)=a,$ so $\mathbf{f(p(x))=p(x)}$." – Subhasis Biswas Sep 11 '19 at 20:47
  • I am not getting why is it. The fixing of $\mathbb{Q}$ is clear to me. – Subhasis Biswas Sep 11 '19 at 20:48
  • @SubhasisBiswas Write $p(x)=\sum a_n x^n.$ Then because each $a_n \in \Bbb Q, \forall x \in F$ we have $f(p(x))=f(\sum a_n x^n)= \sum f(a_n x^n)= \sum f(a_n)f(x^n)= \sum a_nf(x^n)=p(f(x))$. That means that $f$ fixes $p$ as a polynomial over an indeterminate $x$, but you can also see directly that roots of $p$ in $F$ must be in $1-1$ correspondence with roots of $p$ in $K$. – Robert Shore Sep 11 '19 at 20:51
  • This is clear. But in your answer you have yourself written that $f(p(x))=p(x)$. Look again. – Subhasis Biswas Sep 11 '19 at 20:55
  • Viewing $x$ as an indeterminate, what I wrote is correct, because an isomorphism on an abstract polynomial acts only on the coefficients of the polynomial. – Robert Shore Sep 11 '19 at 20:56
  • It is essentially the same, then (notational standpoint)? – Subhasis Biswas Sep 11 '19 at 20:59
  • I think that's right. I edited my comment above to expand on the reasoning, which I hope clears things up a bit. – Robert Shore Sep 11 '19 at 21:01
  • I have myself used almost the exact same argument to arrive at $\phi(p(x))=p(\phi(x))$ in my question. The only unclear part was the notation in your answer. Everything else is very neat. – Subhasis Biswas Sep 11 '19 at 21:04