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Let $R=\mathbb{Z}[x], I=(x^2+1,x+1).$ Prove that $R/I \cong \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}/2\mathbb{Z}$.

I am confused with the rather messy looking of $R/I$. My first step is to define the homorphism $f: R/I \rightarrow \mathbb{Z}[i]$ but then I don't know what to do, please help!

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Find the kernel of the obvious homomorphism ${\bf Z}[x]\to{\bf Z}[i]$, prove the map is surjective, invoke the first isomorphism theorem. Now exhibit a homomorphism ${\bf Z}[i]\to{\bf F}_2$ with kernel $(i+1)$ (the map should send $a+bi$ to the sum of the parities of $a$ and $b$). A different way of proving $R/I\cong{\bf F}_2$ would be to show that $(x^2+1,x+1)=(2,x+1)$ and employing the very useful identity $(R/(a))/(b)\cong R/(a,b)$. – anon May 27 '13 at 9:03
sorry, I still don't quite understand.... Define $\phi : \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ by $\phi (f(x))=f(i)$. Then $\phi$ is a surjective ring homorphism with $\ker \phi= (x^2+1)$. By First Isomorphism Theorem, we have $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$... But then how can I further show that $R/I \cong \mathbb{Z}[i]/(i+1)$...? Sorry for a dumb question... – truvian May 27 '13 at 9:16
The ideal $(i+1)$ in ${\bf Z}[i]$ lifts via third isomorphism theorem to the ideal $(x+1,x^2+1)$ in ${\bf Z}[x]$. That is, ${\bf Z}[x]/(x^2+1,x+1)\cong({\bf Z}[x]/(x^2+1))/(x+1)\cong{\bf Z}[i]/(i+1)\cong{\bf F}_2$. – anon May 27 '13 at 9:19
Why is $(\mathbb{Z}[x]/(x^2+1))/(x+1) \cong \mathbb{Z}[i]/(i+1)$...? I don't understand what we have to do with 3ed isomorphism theorem... – truvian May 27 '13 at 9:34
Nevertheless, thank you for your patience for answering my questions.... – truvian May 27 '13 at 9:34
up vote 2 down vote accepted

Remember the third ismorphism theorem for rings: $\,R\,$ is a ring with ideals $\,I\le J\le R\, $ ,then

$$\left(R/I\right)/\left(J/I\right)\cong R/J$$

In your problem we have

$$\langle x^2+1\,\rangle\le\langle\,x^2+1\,,\,x+1\,\rangle\le\Bbb Z[x]\implies$$

$$\left(\Bbb Z[x]/\langle\,x^2+1\,\rangle\right)/\left(\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\right)\cong\Bbb Z[x]/\langle\,x^2+1\,,\,x+1\,\rangle$$

Now just apply that under the same isomorphism , we have

$$\Bbb Z[x]/\langle\,x^2+1\,\rangle\cong\Bbb Z[i]\;,\;\;\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\cong\langle\,x+1\,\rangle$$

the isomorphism being the one we get from the first isomorphism theorem and the homomorphism

$$\phi:\Bbb Z[x]\to\Bbb Z[i]\;,\;\;\phi(f(x)):=f(i)\;\;\;(\text{i.e., determined by}\;\;x\mapsto i\,)$$

soo that under the above, $\,\Bbb Z[x]\ge\langle\,x+1\,\rangle\mapsto\langle i+1\,\rangle\le\Bbb Z[i]\;$

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Alternatively, one may use the $\color{blue}{\rm FIT}$ = First (vs. Third) Isomorphism Theorem as follows.

$\quad \smash[t]{\Bbb Z\stackrel{h}{\to}}\, \Bbb Z[i\,]/(1\!+\!i)\ \ {\rm is}\ \ \color{#0b0}{\bf onto,}\ \ {\rm by\ \ mod}\,\ 1\!+\!i\!:\ i\,\equiv -1\Rightarrow\:a\!+\!b\,i\,\equiv a\!-\!b\in \Bbb Z\ \\[0.1em] \quad n\in \ker\ h\iff 1\!+\!i\,\mid n\iff \dfrac{n}{1\!+\!i}\, =\, \dfrac{n\,(1\!-\!i\,)}2\,\in\, \Bbb Z[i\,] \iff \color{#c00}2\mid n\\[1.2em] \quad {\rm So} \ \ \ \Bbb Z[i\,]/(1\!+\!i\,)\, \color{#0b0}{\bf =\ Im\:h}\!\smash[t]{\stackrel{\color{blue}{\rm\ \ \ FIT_{\phantom{I^2}}}}\cong} \Bbb Z/\ker h \,=\, \Bbb Z/\color{#c00}2\,\Bbb Z$

Similarly for $\,\Bbb Z[x]/I,\,\ I = (x^2\!+\!1,x\!+\!1).\,$ $\, x\equiv -1\ $ so $\ h\,$ is onto. $\ \ker h = \color{#c00}2\,\Bbb Z\,\ $ by

$\quad \color{#c00}2 = x^2\!+\!1-(x\!-\!1)(x\!+\!1)\in I,\ \ \ n\in I\,\Rightarrow\, n = (x^2\!+\!1)f+(x\!+\!1)g \smash[t]{\stackrel{\large\,\ x\,=\,-1\,\ }\Rightarrow} n = \color{#c00}2f(-1)$

Similarly $\ \Bbb Z[x]/(f(x),x\!-\!a)\,\cong\,\Bbb Z/\color{#c00}{f(a)}\Bbb Z\ \ $ by $\, \ f(x)\equiv \color{#c00}{f(a)}\,\pmod{\!x\!-\!a}$

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