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On this released exam, it asks at 2g (slightly modified wording):

Give a brief example or show there does not exist an ideal $I$, $I \subseteq \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/I$ is isomorphic to $\mathbb{Z}[i]$ the Gaussian integers.

I have a lot of trouble with quotient groups. I have a somewhat decent suspicion that:

$$I = \langle x^2+1 \rangle$$

Is the solution, since the zero of this polynomial is $i$. Can someone help clarify why this is the case if it is, and point me to the right direction if it is not?

Dair
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2 Answers2

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You might for example consider the mapping $$f : \mathbb{Z}[X] \to \mathbb{C}$$ defined by $f(P) = P(i)$, and use some "isomorphism theorem"...

Tlön Uqbar Orbis Tertius
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  • If understand this correctly: $f$ is the evaluation homomorphism, if we restrict the domain of $f : \mathbb{Z}[x] \to \mathbb{Z}[i]$ the same still holds. However, the first isomorphism theorem implies that the quotient group is isomorphic to this subgroup. However, $f$ is also surjective, so the quotient group is isomorphic to $\mathbb{Z}[i]$? – Dair May 04 '15 at 20:09
  • Yes, obviously I could have restricted the domain to $f : \mathbb{Z}[X] \to \mathbb{Z}[i]$, so that $f$ is surjective. By the isomorphism theorem, we have $\mathbb{Z}[X]\big/ \ker (f) \simeq \mathbb{Z}[i]$. All you have to do now is to find that $\ker (f)$ is the ideal generated by $X^2 + 1$ : this should not be too difficult using the properties of polynomials. – Tlön Uqbar Orbis Tertius May 05 '15 at 07:41
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It is useful to recall the very definitions of the rings in question. $\mathbb Z[i]$ is by definition the smallest subring of $\mathbb C$ that contains $\mathbb Z$ and also $i$. Its elements are all obtainable by finitely many ring operations, i.e., adding, subtracting, multiplying of integers and/or $i$. Likewise the ring $\mathbb Z[x]$ has a similar property: All elements are obtainable from integers and $x$ in finitely many steps via adding, subtracting, multiplying. Moreover, $\mathbb Z[x]$ has by definition the univresal property that for any ring $R$, element $r\in R$, ring homomrphism $\phi\colon \mathbb Z\to R$, there exists a unique homomorphism $\Phi\colon\mathbb Z[x]\to R$ such that $\Phi|_{\mathbb Z}=\phi$ and $\Phi(x)=r$.

To construct $\Phi\colon\mathbb Z[x]\to\mathbb Z[i]$ such that $\Phi$ is onto it is therefore quite natural to consider as $\phi\colon\mathbb Z\to\mathbb Z[i]$ the embedding and to pick $i$ as image of $x$. Then let $I$ be the kernel of $\Phi$. Per isomorphism theorem, $\mathbb Z[x]/I$ is isomorphic to the image of $\Phi$. As our choice guaranteed surjectivity, we conclude $\mathbb Z[x]/I\cong \mathbb Z[i]$ as desired.

Hagen von Eitzen
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