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Show that $\Bbb Z[x]/\langle x^2+1\rangle\cong \Bbb Z[i]$.

Take the mapping $\phi :\Bbb Z[x]\to \Bbb Z[i]$ to be $\phi(f(x))=f(i)$ which is definitely a ring homomorphism.

Surely $\phi$ i ssurjective as $a+bi\in \Bbb Z[i]$ we have $f(x)=a+bx\in \Bbb Z[x]$

Then $\ker \phi=\{f(x):f(i)=0\}$ To show that $\ker \phi \cong \langle x^2+1\rangle$.

Now definitely $x^2+1\in \ker \phi\implies \langle x^2+1\rangle\subseteq \ker \phi$

Taking $f(x)\in \ker \phi\implies f(i)=0$.

Unable to show that $f(x)\in \langle x^2+1\rangle$.

Please help.

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    $\begingroup$ You seem to be implicitly using that irreducibles are prime in $\,\Bbb Z[x]\,$ (or unique factorization or related results). That inference should be explicitly justified by citing the Theorem you are invoking. Further you should justify the irreducbility claim. Often such exercises are designed to help you understand how such results do and don't generalize, so it is important to demonstrate precise understanding of such. $\endgroup$
    – Bill Dubuque
    Sep 29, 2017 at 14:33
  • $\begingroup$ @BillDubuque; I only used the fact that we can't write $x^2+1$ as a product of lower degree polynomials in $\Bbb Z[x]$ $\endgroup$
    – Learnmore
    Sep 29, 2017 at 14:39
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    $\begingroup$ I assume you are looking at the homomorphism $\phi \colon {\mathbb Z}[x] \mapsto {\mathbb Z}[i]$ given by $x \mapsto i$, i.e., $f(x) \mapsto f(i)$? The kernel of that map is $\langle x^2 + 1\rangle$. Why do you think it isn't? $\endgroup$
    – Magdiragdag
    Sep 29, 2017 at 15:25
  • $\begingroup$ @Magdiragdag; I explained where my problem is ; $\endgroup$
    – Learnmore
    Sep 29, 2017 at 15:37
  • $\begingroup$ @Magdiragdag;Can you help now $\endgroup$
    – Learnmore
    Sep 29, 2017 at 15:51

1 Answer 1

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Let $f (x)\in \Bbb Z [x] $ such that $f (i) =0$. Since $x^2+1$ is monic there exists $q (x),r (x)\in\Bbb Z [x] $ such that $f (x)=(x^2+1)q (x)+r (x) $ and $r (x)= ax+b $ with $a,b\in \Bbb Z $. Then $0=r (i)=ai+b$, thus $a=b=0$ hence $f (x)\in\langle x^2+1\rangle $.

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  • $\begingroup$ How can you write that?? $\Bbb Z[x]$ is not a ED $\endgroup$
    – Learnmore
    Sep 29, 2017 at 15:37
  • $\begingroup$ @Maths_Student As long as the leading coefficient of the divisor is a unit (as it is in this case since $x^2 +1$ is monic), you can still divide. $\endgroup$
    – Viktor Vaughn
    Sep 29, 2017 at 15:41
  • $\begingroup$ Can't this be done without division algorithm? $\endgroup$
    – Learnmore
    Sep 29, 2017 at 15:50
  • $\begingroup$ Division algorithm holds whenever the divisor has unitary leading coefficient, as in this case. Thus my answer is correct, please remove downvote. $\endgroup$
    – Fabio Lucchini
    Sep 29, 2017 at 15:56

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