# $\mathbb{F_2}[x]/\langle x^2-2\rangle \cong \mathbb{F_2[x]/\langle x^2-3\rangle }$

Prove that the following rings are isomorphic, where $\mathbb{F_2} = \mathbb{Z}/2\mathbb{Z}$

$$\mathbb{F_2}[x]/\langle x^2-2\rangle \cong \mathbb{F_2[x]/\langle x^2-3\rangle }$$

My attempt:

By checking all elements of the field $\mathbb{F_2}$ we can conclude that both $x^2-2$ and $x^2-3$ are reducibles.

Could you point me towards the direction of the correct proof.

• Are your quotients $\mathbb{F}_2[x]$ and not $\mathbb{F}_2$? Also, in $\mathbb{F}_2$, $2=0$ and $3=1$, so, do you mean $\mathbb{F}_2[x]/\langle x^2\rangle\simeq\mathbb{F}_2[x]/\langle x^2+1\rangle$?. – Michael Burr Mar 2 '17 at 15:35
• @Micheal Burr ...sorry i have edited it....Yes thats what i mean – spaceman_spiff Mar 2 '17 at 15:44

$x^2-2\equiv x^2\pmod {2}$, so it is reducible. Also note $x^2-3\equiv x^2+1\equiv (x+1)^2\pmod{2}$.
Then you can take the map from $F_2[x]\to F_2[x]/(x+1)^2$ that sends $x\mapsto x+1$, and confirm it gives rise to an isomorphism of $F_2[x]/(x^2)$ with $F_2[x]/(x+1)^2$ via the first isomorphism theorem.
• Well, you've got an element in the left ring that squares to zero ($x$) and an element that squares in the right ring to zero ($x+1$) so it makes a lot of sense to map one to the other using the universal property of polynomial rings and the first isomorphism theorem. That way you don't have to prove anything about it being a homomorphism. – rschwieb Mar 2 '17 at 16:02