2
$\begingroup$

Let $R = \mathbb{Z}[ i ] / (5)$ .

How should I prove that $5 = (2+i) (2-i)$ is a prime factorization in $\mathbb{Z}[i]$? Can we deduce from this that R is not an integral domain? How?

I know that we can prove any ideal in R is principal.

Now I want to prove the classification theorem for modules over $R$ :

There exist modules $M_1, M_2$ such that any finitely generated module $M$ over $R$ is isomorphic to the direct sum $M_1^r \oplus M_2^s$, where $M_1^r$ is the direct sum of $r$ copies of module $M_1$, and similarly for $M_2$.

I notice that $R$ is not an PID...........

Do you have any ideas how to prove this?

$\endgroup$
4
  • 3
    $\begingroup$ What have you tried so far? Do you know about the Chinese Remainder Theorem in (say) a PID? This tells you something about the structure of $R/(p_1 p_2)$ where $p_1$ and $p_2$ are nonassociate prime elements. $\endgroup$ Commented Dec 13, 2012 at 19:38
  • $\begingroup$ What is the Chinese Remainder Theorem in a PID? $\endgroup$ Commented Dec 13, 2012 at 19:59
  • $\begingroup$ Wikipedia might be a good start. $\endgroup$ Commented Dec 14, 2012 at 9:17
  • $\begingroup$ Why did you post the same question splitted in two? Duplicate of math.stackexchange.com/questions/258421/… and math.stackexchange.com/questions/258523/… $\endgroup$ Commented Dec 14, 2012 at 19:45

4 Answers 4

Reset to default
5
$\begingroup$

You don't need to know that $5 = (2-i)(2+i)$ is a prime factorization; all you need is that the two factors are not units.

One way to see that $2-i$ is not a unit is by computation:

$$ \mathbb{Z}[i] / (2-i) \xrightarrow{i \to x} \mathbb{Z}[x] / (x^2 + 1, 2-x) \xrightarrow{x \to 2} \mathbb{Z} / (2^2 + 1) \cong \mathbb{Z} / 5\Bbb Z $$

(all arrows are isomorphisms). The result isn't the zero ring, so $2-i$ is not a unit. The fact the result is a domain does additionally prove that $2-i$ is prime, though.

$\endgroup$
3
  • $\begingroup$ Alternatively, $N(2\pm i)=5\neq1$ so $2\pm i$ are not units. $\endgroup$ Commented yesterday
  • $\begingroup$ @19021605 That was not a typo and should not have been edited. Again, please refrain from making trivial edits to old posts. $\endgroup$ Commented yesterday
  • $\begingroup$ @BillDubuque I apologize. I didn't know $\Bbb Z/ n$ is a valid notation. $\endgroup$ Commented yesterday
2
$\begingroup$

If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irreducible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

$\endgroup$
7
  • $\begingroup$ What are the ideals of R = Z[i] / (5)? $\endgroup$ Commented Dec 13, 2012 at 20:15
  • $\begingroup$ The ideals of $R$ are precisely those of the form $I/\langle 5\rangle$, where $I$ is an ideal of $\Bbb Z[i]$ containing $(5)$. This is a consequence of what is sometimes referred to as the Correspondence Theorem. Since $\Bbb Z[i]$ is a PID, then the ideals of $\Bbb Z[i]$ that contain $\langle 5\rangle$ are precisely: $\langle 5\rangle$, $\langle 2+i\rangle$, $\langle 2-i\rangle$, and $\langle 1\rangle=\Bbb Z[i]$. Hence, $R$ has only $4$ ideals: $\langle 0\rangle$, $\langle 2+i\rangle/\langle 5\rangle$, $\langle 2-i\rangle/\langle 5\rangle$, and $R$. $\endgroup$ Commented Dec 13, 2012 at 20:24
  • $\begingroup$ Are they prime ideals? $\endgroup$ Commented Dec 13, 2012 at 20:28
  • $\begingroup$ You should be able to prove that each ideal of $R$ is principal, and using the result from this related answer in conjunction with Third Isomorphism Theorem, you should be able to prove that both non-trivial ideals of $R$ are prime (maximal) ideals, since $\Bbb Z/\langle 5\rangle$ is an integral domain (a field). $\endgroup$ Commented Dec 13, 2012 at 20:31
  • $\begingroup$ For modules over R=Z[i]/(5) , I am trying to prove the so-called classification theorem: There should exist modules M_1 , M_2 such that any finitely generated module M over R is isomorphic to the direct sum M_1^r ⊕ M_2^s , where M_1^r is the direct sum of r copies of module M_1, and similarly for M_2 . Do you have any ideas how to do this? $\endgroup$ Commented Dec 13, 2012 at 20:38
1
$\begingroup$

$\mathbb{Z}[i]$ is a Euclidian domain, so in particular it is also a UFD. We know in UFDs that an element is prime if and only if it is irreducible. If you pull out your trusty copy of Dummit and Foote, you can see on p. 291 that the list of irreducibles in the Gaussian integers includes:

1) $1+i$ and $1-i$

2) Prime integers such that $p \equiv 3\mod 4$

3) If $p \equiv 1\mod 4$ then $p$ can be uniquely expressed (up to signs of $a$ and $b$, and interchanging) as $p = a^2+b^2$ for integers $a$ and $b$, and the elements $a+ib$ and $a-ib$ are irreducible.

An ideal $I$ in a ring $R$ is prime if and only if $R/I$ is an integral domain. You have already noticed that $5$ is not an irreducible element, so $(5)$ is not prime, therefore $\mathbb{Z}[i]/(5)$ is not an integral domain.

Edit: Just noticed you had asked other questions too. Getting to them now.

-What are the ideals in $R$: They're in bijective correspondence with ideals in $\mathbb{Z}[i]$ containing the ideal $(5)$. These should be generated by elements which divide $5$ in $\mathbb{Z}[i]$

-Every ideal of $R$ is principal. See this answer https://math.stackexchange.com/a/90014/29076

$\endgroup$
4
  • $\begingroup$ What are the ideals of R = Z[i] / (5)? ?? $\endgroup$ Commented Dec 13, 2012 at 20:16
  • $\begingroup$ The ideals of $\mathbb Z[i]/(5)$ are the ideals of $Z[i]$ that contain $(5)$. In particular $\mathbb Z[i]$ is a pid so you can characterize these ideals by the prime factorization of $5$ in the Gaussian integers. $\endgroup$ Commented Dec 13, 2012 at 20:20
  • $\begingroup$ Are they all prime ideals? $\endgroup$ Commented Dec 13, 2012 at 20:23
  • $\begingroup$ For modules over R=Z[i]/(5) , I am trying to prove the so-called classification theorem: There should exist modules M_1 , M_2 such that any finitely generated module M over R is isomorphic to the direct sum M_1^r ⊕ M_2^s , where M_1^r is the direct sum of r copies of module M_1, and similarly for M_2 . Do you have any ideas how to do this? $\endgroup$ Commented Dec 13, 2012 at 20:39
0
$\begingroup$

To prove that $R$ is not an integral domain all we need are that $2-i$ and $2+i$ are nonzero. For $\alpha \in \mathbb{Z}[i]$ to be zero in $R$ would mean that $\alpha \in (5)$, so there exists $a + bi \in \mathbb{Z}[i]$ such that $5(a+bi) = 5a + (5b)i =\alpha$. Clearly neither $2-i$ or $2+i$ are of this form, thus $R$ is not an integral domain.

If you did in fact want to prove that $2-i$ and $2+i$ are primes in $\mathbb{Z}[i]$, consider the following: If $2-i$ was not prime we would be able to write $2-i = \alpha \beta$ for some $\alpha, \beta \in \mathbb{Z}[i]$ that are nonunits. Multiplying both sides by their conjugates we get that \begin{equation} (2-i)(2+i) = 5 = (\alpha\bar\alpha)(\beta\bar\beta). \end{equation} Note that $\alpha\bar\alpha$ and $\beta\bar\beta$ are both integers. Then since $5$ is prime we have that one of $\alpha\bar\alpha$ or $\beta\bar\beta$ must be $\pm 1$. Without loss of generality, if $\alpha\bar\alpha = \pm 1$ then $\alpha$ is a unit, proving that $2-i$ cannot be written as a product of two nonunits and is thus prime. The same process works for $2+i$.

New contributor
Grant Hacking is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.