# Two finite fields are isomorphic. [duplicate]

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.

• 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

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 $F[X]/(f)\simeq F[X]/(g)$. – user26857 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.) – user26857 Feb 27 '13 at 23:09

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.

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 $\rm\:f.\:$