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

Question

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?

  2. 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 $R[x-a] \cong R[x-b]$?

Answer 1

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.

Answer 2

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 ${\Bbb F}_2[x]/\langle x^2+1\rangle$: $$x\cdot x = x^2=1,\quad x\cdot(x+1)=x^2+x=x+1,\quad (x+1)(x+1)=0.$$