TITLE: Prove that the fields $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4 \rangle$ are isomorphic
QUESTION [14 upvotes]: I have been stuck in this problem for some time now.

Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$
  and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic, each
  having $121$ elements.

The first part is easy to prove since there is no element of $\mathbb Z_{11}$ that satisfies either of the polynomials given in the question. However, proving that  $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic has been a challenge for me. How do I proceed? Moreover, how do I show that the the fields $\mathbb {Z}_{11}[x]/\langle x^2+1\rangle$ and $\mathbb {Z}_{11}[x]/\langle x^2+x+4\rangle$ each have $121$ elements?

REPLY [6 votes]: Hint $\ $ You seek an isomorphism of $\:\mathbb F_{11}[\sqrt{-1}]\:$ and $\rm\:\mathbb F_{11}[5-\sqrt{-1}\:\!],\:$ since $\rm\:x^2+x+4\:$ has roots $$\rm\dfrac{-1\pm\sqrt{-15}}2 \:\equiv\: \dfrac{10\pm\sqrt{-4}}2\:\equiv\: 5\pm\sqrt{-1}\pmod{11}$$
The discriminant (mod squares) characterizes isomorphism classes of quadratic extensions.