# Questions on a proof that $\mathbb F_2[x]/(x^2-1)$ and $\mathbb F_2[x]/(x^2)$ are isomorphic.

This has been proven more generally for a field of characteristic 2 in another question.

Here is the proof from brianbi.ca $$\mathbb F_2[x]/(x^2) \cong \mathbb F_2[x+1]/((x + 1)^2) = \mathbb F_2[x]/(x^2 + 2x + 1) = \mathbb F_2[x]/(x^2 - 1)$$

1. Why is $$\mathbb F_2[x]/(x^2) \cong \mathbb F_2[x+1]/((x + 1)^2)$$?

-We can use the correspondence theorem by the surjective homomorphism $$\varphi:\mathbb F_2[x] \to \mathbb F_2[x+1], \varphi(p(x))=p(x+1)$$, so the ideal $$(x^2)$$ corresponds to the ideal $$(x + 1)^2$$. Is this correct?

• Is injectivity irrelevant?

1. Why is $$\mathbb F_2[x+1]/((x + 1)^2) = \mathbb F_2[x]/(x^2 + 2x + 1)$$

?

• $$\mathbb F_2[x+1] \cong \mathbb F_2[x]$$ by the same homomorphism $$\varphi$$ because $$\varphi$$ is injective too. Is this correct? I think this proves only $$\cong$$ and not necessarily $$=$$.

• Do we actually have $$\mathbb F_2[x+1] = \mathbb F_2[x]$$? If so, how? In general for a ring $$R$$ and $$a,b \in R$$, when is $$R[x-a] = R[x-b]$$? When is $$R[x-a] \cong R[x-b]$$?

I seems to me that in fact $$\mathbb{F}_2[x+1] = \mathbb{F}_2[x]$$, because we have $$x+1 \in \mathbb{F}_2[x]$$, and also in characteristic 2 we have $$2=0$$, hence $$x = x+2 = (x+1) + 1 \in \mathbb{F}_2[x+1]$$.
More generally polynomials in $$x$$ with coefficients in any field $$\mathbb{F}$$ are the same as polynomials in $$x+1$$ with coefficients in $$\mathbb{F}$$, since we can always write $$x = (x+1) - 1$$, so $$p(x) = p((x+1)-1) = q(x+1)$$ for some polynomial $$q$$, and we can pass from $$q(x+1)$$ to $$p(x)$$ by writing $$q(x+1)$$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.
The quotient rings $${\Bbb F}_2[x]/\langle x^2+1\rangle$$ and $${\Bbb F}_2[x]/\langle x^2\rangle$$ are isomorphic as $${\Bbb F}_2$$-vector spaces.
In view of multiplication, it is obvious that a ring isomorphism $$\phi:{\Bbb F}_2[x]/\langle x^2\rangle\rightarrow {\Bbb F}_2[x]/\langle x^2+1\rangle$$ maps $$x$$ to $$x+1$$.
Indeed, check that in $${\Bbb F}_2[x]/\langle x^2\rangle$$: $$x\cdot x = 0,\quad x\cdot(x+1)=x^2+x=x, \quad(x+1)(x+1)=x^2+1=1,$$ while in $${\Bbb F}_2[x]/\langle x^2+1\rangle$$: $$x\cdot x = x^2=1,\quad x\cdot(x+1)=x^2+x=x+1,\quad (x+1)(x+1)=0.$$