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My attempt $:$

If $\alpha$ is algebraic over a field $F$ then by first isomorphism theorem it is clear that $$F [x] / (f (x)) \simeq F [\alpha]$$ where $f (x)$ is the irreducible polynomial for $\alpha$ over $F$.Now since $F [x]$ is a PID and $f (x)$ is irreducible over $F$ so $(f (x))$ is a maximal ideal and hence $F [x]/ (f (x))$ is a field and hence so is $F [\alpha]$.But then $F [\alpha] [y] = F [\alpha,y]$ is a PID. So again the application of first isomorphism theorem yields $$F [\alpha,y] / (g (y)) \simeq F [\alpha,\beta]$$ where $g (y)$ is the irreducible polynomial for $\beta$ over $F [\alpha]$.But again by similar argunment mentioned above we have $F [\alpha,\beta]$ is a field.

Is my reasoning correct at all? Please check it.

Thank you in advance.

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    $\begingroup$ Completely correct. +1 $\endgroup$ – DonAntonio Mar 31 '18 at 17:23
  • $\begingroup$ Thanks @DonAntonio for your verification. $\endgroup$ – Arnab Chatterjee. Mar 31 '18 at 17:25
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    $\begingroup$ @Ar Isn't that exactly what you just proved at the end of your post ? Perhaps though it'd be better to write $$F(\alpha)[y]/(g(t))\cong F[\alpha,\beta]\;,\;\;g(t)\in F(\alpha)[t]\;\;\text{minimal for}\;\;\beta$$ $\endgroup$ – DonAntonio Mar 31 '18 at 17:43
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    $\begingroup$ @DonAntonio Why don't we need an additional hypothesis here that $F[\alpha, \beta]$ is contained in some field. I thought $\alpha = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ and $\beta = \begin{pmatrix} c & -d \\ d & c-d \end{pmatrix}$ with coefficients in $\Bbb{Q}$ would provide a counter-example. $\alpha$ satisfies the equation $x^2-2ax+a^2+b^2$ and $\beta$ satisfies the equation $x^2-(2c-d)x+c^2-cd+d^2$, so $\Bbb{Q}[\alpha]$ and $\Bbb{Q}[\beta]$ are both fields. But $\Bbb{Q}[\alpha, \beta]$ is isomorphic to the ring of 2 x 2 matrices over $\Bbb{Q}$ and certainly not a field. $\endgroup$ – sharding4 Mar 31 '18 at 18:50
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    $\begingroup$ @sharding4 Who said we don't? Though it is trivially contained in the functions field $\;F(\alpha,\beta)\;$ ...Not to mention that your example uses polynomials valuated at matrices: that is messy, in general. $\endgroup$ – DonAntonio Mar 31 '18 at 19:51

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