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Describe the maximal ideals in the ring of Gaussian Integers $\Bbb Z[i]$.

So first of all my question would be - Is it possible to write any ideal of $\Bbb Z[i]$ (which happens to be a PID) as $\langle a\rangle$ or does it have to be $\langle a + bi\rangle$ or are they the same thing?

i.e Can ANY Ideal in $Z[i]$ be written as $\langle a\rangle$ ?

Further, I am not able to continue. Please help!

Some possible references (I looked through these but to no avail):

  1. Maximal ideals in the ring of Gaussian integers

  2. How to find all the maximal ideals of $\mathbb Z_n?$

  3. Ideals in the ring of Gaussian integers
AJ_
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  • Use the definition of PID. and an Ideal – Kumar Jul 14 '19 at 11:41
  • @Kumar Could not conclude anything substantial from the above link. – AJ_ Jul 14 '19 at 11:54
  • What do you denote $a$ in $\langle a\rangle$? An ordinary integer or a Gauß integer? – Bernard Jul 14 '19 at 11:58
  • I think if you look at the question and answer of the link, you would be able to get the answer. Anyways. https://math.stackexchange.com/questions/764437/the-gaussian-integers-are-isomorphic-to-mathbbzx-x21?noredirect=1&lq=1 This link will help you to understand more if you understand polynomial rings. I would also recommend you to have a look at Dummit and Foote, Abstract Algebra. Happy idealizing. :P – Kumar Jul 14 '19 at 12:04
  • Well, $;\Bbb Z[i];$ is an Euclidean domain and is thus a PID, so yes: every ideal in the ring can be written as $;\langle a\rangle;$ ...as long as $;a=x+iy,,,,x,y\in\Bbb Z;$ . – DonAntonio Jul 14 '19 at 12:05
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    The maximal ideals of $\Bbb{Z}[i]$ are those generated by the primes of this ring. Whether a rational (=usual) prime $p$ remains a prime in $\Bbb{Z}[i]$ depends on its residue class modulo $4$. If $p\not\equiv-1\pmod4$, then it factors further. – Jyrki Lahtonen Jul 14 '19 at 17:15

2 Answers2

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No; not every ideal of $\Bbb{Z}[i]$ is of the form $\langle a\rangle$ for some integer $a$. A simple example is the ideal $\langle 1+i\rangle$. It contains the integer $2=(1+i)(1-i)$, so if it is generated by an integer, then it must be generated by a divisor of $2$. But then it should equal either $\langle1\rangle=\Bbb{Z}[i]$ or $\langle2\rangle$, which it doesn't.

Servaes
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Since $\mathbf Z[i]$ is a P.I.D. – actually a Euclidean domain with the norm: $N(a+bi)=a^2+b^2$ as a Euclidean function, maximal ideals are generated by irreducible elements.

Also $a+bi$ is irreducible if $N(a+bi)$ is prime, because the norm is multiplicative, i.e. $$N\bigl((a+bi)(c+di)\bigr)=N(a+bi)N(c+di).$$

Bernard
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