Justifying $\,\Bbb{Z}[i]/\langle 2\rangle\cong \Bbb{Z}[X]/\langle X^2+1,2\rangle\cong \Bbb{F}_2[X]/\langle X^2+1\rangle$

I am asked to describe the quotient ring $\,\Bbb{Z}[X]/\langle X^2+1,2\rangle$ in two different ways.

I would say that $\Bbb{Z}[X]/\langle X^2+1,2\rangle$ is isomorphic to $\Bbb{F}_2[X]/\langle X^2+1\rangle$ and $\Bbb{Z}[i]/\langle 2\rangle$ but I am not sure how can I justify that. Does we have $\langle X^2+1,2\rangle=\langle X^2+1\rangle\langle2\rangle$? I think the answer is no, because $\langle X^2+1,2\rangle=2P(X)+(X^2+1)Q(X)$.

How can I justify the two isomorphism ?

Solution 1:

This quotient $\rm \color{#c00}{reci}\color{#0a0}{procity}$ arises by applying Third Isomorphism Theorem in two different ways

$\begin{align} (R/J)\,/\,(I\!+\!J)/J\ &\cong \ R/(I\!+\!J)\ \cong\ (R/I)\,/\,(J\!+\!I)/I\\[.1em] \text{Or, abusing notation:}\, \ \ \ \ (R/\color{#0a0}J)/\color{#c00}I\, &\cong \ R/(I\!+\!J)\ \cong\,(R/\color{#c00}I)/\color{#0a0}J^{\phantom{1^{1^1}}} [\rm \color{#c00}{reci}\color{#0a0}{procity}]\\[.4em] \text{OP is s special case:}\,\ R=\Bbb Z[x],\,\ I &=(2),\,\ J = (x^2\!+\!1) =: (f),\,\ \rm explicitly\\[.4em] \Bbb Z[\:\!i\:\!]/2 \,&\cong\, (\Bbb Z[x]/f)\,/\,(2,f)/f\\[.2em] &\cong\, \ \Bbb Z[x]/(2,f)\\[.2em] &\cong\, (\Bbb Z[x]/2)\,/\,(f,2)/2\\[.2em] &\cong\ \ \Bbb F_2[x]/f \quad \end{align}$

Remark $ $ While the abused notation better highlights the innate reciprocity, as for any notational abuse, one should be certain to clearly understand the more precise rigorous denotation in order to avoid being led astray by the less rigorous notation.

Note $\,x^2+1 = (x+1)^2\,$ in $\Bbb F_2\,$ so $\,\Bbb Z[i]/2\cong \Bbb F_2[x]/(x+1)^2\cong \Bbb F_2[t]/t^2.\,$ The ring $\,R[t]/t^2\,$ is known as the algebra of dual numbers over the ring $\,R.\,$ Such rings and higher order analogs $\, R[t]/t^n \;$ prove quite useful when studying (higher) derivations algebraically, since they provide very convenient algebraic models of tangent / jet spaces. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. See this post for further discussion and links.

Solution 2:

Use the first isomorphism theorem.

Define $\varphi:\mathbb{Z}[x] \to \mathbb{Z}[i]/(2)$ by $\varphi(f(x))=f(i)+(2)$.

It's not hard to see that $\varphi$ is an onto homomorphism. If you establish that its kernel is $(x^2+1,2)$, you'll have one of your isomorphisms (using the first isomorphism theorem).

Next, define $\psi:\mathbb{Z}[x] \to \mathbb{Z}_2[x]/(x^2+1)$ by $\psi(f(x))=\bar{f}(x)+(x^2+1)$ where $\bar{f}(x)$ is $f(x)$ with its coefficients reduced mod $2$.

Again, it's not hard to see that $\psi$ is an onto homomorphism. Then again establish it has the kernel $(x^2+1,2)$ and done!