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