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Does there exist any isomorphism between the fields $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 3]$ ?

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    $\begingroup$ Does the first field contain an element whose square is $3$? $\endgroup$ – André Nicolas Jul 11 '16 at 18:30
  • $\begingroup$ No... there is no such element in the first field@Andre $\endgroup$ – BijanDatta Jul 11 '16 at 18:33
  • $\begingroup$ Well, that settles it, for if an isomorphism takes $a+b\sqrt{2}$ to $\sqrt{3}$, then $(a+b\sqrt{2})^2=3$. $\endgroup$ – André Nicolas Jul 11 '16 at 18:39
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Suppose there were an isomorphism of fields $\phi: \mathbb{Q}[ \sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{3}]$. It is straightforward to show that such an isomorphism would necessarily fix $\mathbb{Q}$ since it would fix $1$ (see here).

Next, we must have $\phi(\sqrt{2}) = a + b\sqrt{3}$ for some $a, b \in \mathbb{Q}$. The multiplicativity of $\phi$ gives us $\phi( \sqrt{2}) \phi(\sqrt{2}) = \phi \Big( \left(\sqrt{2} \right)^2 \Big) = \phi(2) = a^2 + 2ab \sqrt{3} + 3b^2$. Because $\phi$ fixes $\mathbb{Q}$, we arrive at $a^2 + 2ab \sqrt{3} + 3b^2 = 2$, and herein lies a contradiction.


$\underline{\textbf{Caution}}$: Although $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are not isomorphic as fields, they are isomorphic as vector spaces since they are both of dimension $2$ over $\mathbb{Q}$. So without context, one needs to be careful with unqualified statements like "$\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are not isomorphic." Click here for further discussion.

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  • $\begingroup$ I cannot understand the statement "an isomorphism would necessarily fix $Q$". May you explain plz. $\endgroup$ – BijanDatta Jul 11 '16 at 18:46
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    $\begingroup$ @B.R.Datt: Start with the insight that an isomorphism necessarily fixes $1$ as multiplicative identity. It follows that an isomorphism fixes $\mathbb{N}$ and consequently all of $\mathbb{Q}$. $\endgroup$ – hardmath Jul 11 '16 at 19:40
  • $\begingroup$ Added some cross-references @BijanDatta $\endgroup$ – Kaj Hansen Apr 11 at 8:47
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Consider the equation $x^2=2$ which has a solution in the first field. Suppose $f \colon \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{3}]$ is an isomorphism, so $f(x)^2 = f(x^2) = f(2)$. But note that $f(2)=2$, as an isomorphism must fix the multiplicative identity. Hence, we have $y^2=2$ where $y = f(x) \in \mathbb{Q}[\sqrt{3}]$. Then $y^2 = (a+b\sqrt{3})^2 = 2$ and so $a^2+2ab\sqrt{3} +3b^2=2$. By a rationality argument then $a^2+3b^2 =2$ and so $a$ or $b$ is zero. If $b$ is zero then we conclude that $2$ has a rational square root. If $a$ is zero then we conclude that $2/3$ has a rational square root. Both lead to contradictions.

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  • $\begingroup$ What ia the relation between two isomorphic fields and roots of the polynomials? I need more explanation. #DanRust $\endgroup$ – BijanDatta Jul 11 '16 at 18:44
  • $\begingroup$ I think you mean "By a rationality argument then $a^2 + 3b^2 = 2$ and $2ab = 0$, and so $a$ or $b$ is zero? Of course, the fact that either $a=0$ or $b=0$ follows from $2ab=0$, not $a^2 + 3b^2 = 2$. $\endgroup$ – MCT Jul 11 '16 at 18:49

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