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Find an ideal $I$ of $\mathbb{Z}[i]$ such that $\mathbb{Z}[i]/{I}$ is a field.

How can one justify the answer in the shortest number of lines?

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You need to when you can write a prime as a sum of squares. –  Baby Dragon Aug 12 '12 at 3:19
    
What is R in the question? –  Abhishek Parab Aug 12 '12 at 3:25

5 Answers 5

If $R$ is a ring and $I$ is a maximal ideal of $R$, then $R / I$ is a field.

$\mathbb{Z}[i]$ is a Euclidean domain, so prime ideals are maximal. Since $\mathbb{Z}[i]$ is a Euclidean domain, it is a PID. Thus it suffices to find the prime elements of $\mathbb{Z}[i]$.

A well-known result is that:

The prime elements of $\mathbb{Z} [i]$ are, up to associates:

(1) 1 + i

(2) p where $p \in \mathbb{Z}$ is prime and $p \equiv 3 \text{ mod } 4$.

(3) $a + bi$ where $a^2 + b^2 = p$ and $p \in \mathbb{Z}$ is prime and $p \equiv 1\text{ mod } 4$.

If $\pi$ is any prime (i.e. an element associate to element of type above), then $\mathbb{Z}[i] / (\pi)$ is a field.

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Write the Gaussian integers as $\Bbb{Z}[x]/(x^2 +1)$. Then notice that for a prime $p$ such that $x^2 +1$ is irreducible mod $p$, you have that $(x^2 + 1,p)$ is a maximal ideal in $\Bbb{Z}[x]$. You will then have

$$\Bbb{Z}[i]/(p) \cong \Bbb{Z}[x]/(x^2 + 1,p) \cong \Bbb{Z}/p\Bbb{Z}[x]/(\overline{x}^2 +1).$$

Now in the right most guy you have $\Bbb{Z}/p\Bbb{Z}[x]$ that is a polynomial ring over a field and by choice $\overline{x}^2 +1$ is irreducible mod $p$ and so this is a field. It follows that $\Bbb{Z}[i]/(p)$ is a field.

Now of course one could ask for the existence of such a prime $p$. It may be the case that $x^2 + 1$ is reducible for every prime $p$ (such polynomials do exist) but we can choose $p=3$ in this case and it would work.

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+1 That was nice and simple –  DonAntonio Aug 12 '12 at 4:30
    
@DonAntonio No need to invoke any algebraic number theory :D –  user38268 Aug 12 '12 at 4:33

Hint $\ \Bbb Z[i]/(i\!-\!1)\cong \Bbb Z/2\:$ is the unique extension of parity from $\,\Bbb Z,\:$ given by defining $\,i\,$ to be odd.

Remark $\ $ Any ring having $\,\Bbb Z/2\,$ as an image has parity structure induced by such. Generally it not be unique, e.g. see this answer for further discussion of rings with parity.

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Let $I=(3)$ then $Z[i]/I=Z_3[i]$ is a field (since $x^2+1$ is irreducible over $Z_3$) having 9 elements.

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$I$ must clearly be a prime ideal, since $R/I$ must be a field. Every prime ideal of $\mathbb Z[i]$ is maximal since its a PID. So $I$ can be taken to be any nonzero prime ideal.

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