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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 ?

Bill Dubuque
  • 272,048
  • Hint: What is the image of $\langle X^2+1,2\rangle$ in $\mathbb{F}_2[X]$ (via the first isomorphism theorem)? – Michael Burr Mar 31 '15 at 15:47
  • Please don't write $Z[X]/<X^2+1,2>$ if you mean $Z[X]/\langle X^2+1,2\rangle$. (I changed it.) Also, you might consider whether you want $\mathbb Z[X]/\langle X^2+1,2\rangle$. ${}\qquad{}$ – Michael Hardy Mar 31 '15 at 16:22
  • @MichaelHardy Some problem with my computer, thanks. –  Mar 31 '15 at 16:34

2 Answers2

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

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

Above, we implicitly employed $R/K \cong h(R)/h(K)\,$ for injective ring hom $h$.

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 $R.\,$ 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, cf. section 8.15 in Jacobson, Basic Algebra II. See also this post for further discussion and links.

Bill Dubuque
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1

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!

Bill Cook
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