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Refering to this question suppose I have $l(x):=x^3+x+1$ and $m(x):=x^3+x^2+1$. Then prove there is an isomorphism between $\mathbb{F}_3 [x]/l(x)$ and $\mathbb{F}_3[x]/m(x)$

I can say that elements for both the fields are same.

$$x^3+x+1= 0,\quad 1 x,\quad x+1,\quad x^2,\quad x^2+1,\quad x^2+x,\quad x^2+x+1$$

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    $\begingroup$ The polynomials are reducible (both have $1$ as a root). Therefore the quotients aren't fields. $\endgroup$ – Ayman Hourieh Apr 21 '14 at 18:43
  • $\begingroup$ If you will look at the question that you cite, you will see that the polynomials in question are $x^3-x+1$ and $x^3-x^2+1$, and if you read my answer there, you will see how to construct the isomorphism. And no, you cannot assert that some elements are the same the way you do. $\endgroup$ – Dilip Sarwate Apr 21 '14 at 18:49
  • $\begingroup$ But I am confused in (a$β^2$+bβ+c)3=a$β^2$+bβ+c−1. How R.H.S is produced? $\endgroup$ – user120838 Apr 21 '14 at 19:06
  • $\begingroup$ Are you sure you aren't looking at these over $\Bbb{F}_2$? In that case the two polynomials would actually be irreducible. For the actual question Dilip's answer in the linked thread is the way to go. $\endgroup$ – Jyrki Lahtonen Apr 21 '14 at 20:49
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If you already know that both are fields then all you have to do is count their order, because finite fields are uniquely determined up to isomorphism by the number of elements they contain.

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    $\begingroup$ The polynomials are reducible (both have $1$ as a root). Therefore the quotients aren't fields. $\endgroup$ – Ayman Hourieh Apr 21 '14 at 18:41

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