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Let $F = \Bbb{Z}_2$. Given the irreducible polynomials $f(x)= x^3 + x + 1$, and $g(y) = y^3 + y^2 + 1$, form the fields $K = F[x]/(f(x))$ and $E = F[y] / (g(y))$. These are fields of order 8 (given), so they must be isomorphic. Is the map $[x] \mapsto [y + 1]$ an isomorphism? It's clearly onto, and and it's one-one since $F$ and $E$ both have 8 elements.

user64276
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  • If it is bijective, all you must do is show that it is a homomorphism, as a field isomorphism is just a bijective homomorphism. – Emily Feb 27 '13 at 22:10

3 Answers3

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Take the $F$-isomorphism $\varphi:F[X]\to F[Y]$ which sends $X$ to $Y+1$. Then $\varphi(f)=g$ and therefore $F[X]/(f)\simeq F[Y]/(g)$.

  • Is your last step a consequence of some isomorphism theorem? – user64276 Feb 27 '13 at 22:23
  • As a general note this method won't always work. See here for more details. – JSchlather Feb 27 '13 at 22:37
  • @JSchlather Actually I don't get your comment. That problem you pointed out deals with the converse of the present question. This one is generally valid: if $\varphi$ is an $F$-automorphism of $F[X]$, $f\in F[X]$ an irreducible polynomial and $g=\varphi(f)$ then $F[X]/(f)\simeq F[X]/(g)$. –  Feb 27 '13 at 22:46
  • @YACP I'm saying that if you know $F[X]/(f) \cong F[X]/(g)$ you can't always find an $F$-isomorphism $\varphi$ of $F[x]$ such that $\varphi(f)=g$. I was trying to say that in regards to showing two such fields were isomorphic, you couldn't always find an $F$-isomorphism. Not that an $F$-isomorphism didn't descend to an isomorphism of fields. I agree that the phrasing was unclear, sorry. – JSchlather Feb 27 '13 at 22:49
  • @JSchlather This question can be used in some sense to give a very simple counter-example to yours. There are (obviously) only two $F$-automorphisms of $F[X]$: $X\mapsto X$ or $X\mapsto X+1$. Now take $f=X^4+X+1$ and $g=X^4+X^2+1$. (I hope I've understood correctly your question.) –  Feb 27 '13 at 23:09
  • @YACP Yes. And actually I had been wondering looking at that question again if you could find one where the polynomials weren't associate after an automorphism was applied to one. Which that example shows you can do as well. – JSchlather Feb 28 '13 at 05:57
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An element of $K$ has a unique representation as $p + (f)$, where $\deg p \leq 2$ and an element of $E$ has a unique representation as $q + (g)$, where $\deg q\leq 2$. Take $A = ax^2 + bx + c + (f)$, $B = a'x^2 + b'x + c + (f)$. \begin{align*} \phi(A + B) &= \phi(ax^2 + bx + c + (f) + a'x^2 + b'x + c + (f))\\ &= \phi((a + a')x^2 + (b + b')x + c + c' + (f))\\ &= (a + a')(y + 1)^2 + (b + b')(y + 1) + c + c' + (g)\\ &= (ay^2 + a + by + b + c + (g)) + (a'y^2 + a' + b'y + b' + c' + (g))\\ &= \phi(A) + \phi(B) \end{align*} Then if you can show $\phi(AB) = \phi(A)\phi(B)$ using a similar calculation, $\phi$ is a bijective homomorphism, and hence an isomorphism.

Stahl
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Hint $\rm\,\ g(x)\, =\, x^3 f(1/x) $

Remark $\ $ If $\rm\:f(x)\:$ has degree $\rm\,n,\:$ then $\rm\: x^n f(1/x)\:$ is the reciprocal polynomial of $\rm\:x\:,\:$ so-named because its roots are reciprocals of the roots of $\rm\:f.\:$ As here, it is easily recognizable since it coefficients are the reverse of that of $\rm\:f.\:$

Math Gems
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