Sanity check on factorization of $\langle 5 \rangle$ in $\mathbb{Z}_{10}$

Question

Looking at $\mathbb{Z}_{10}$, consisting of $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ with addition and multiplication adjusted so the ring is closed under both operations.

Clearly $5 = 3 \times 5 = 5 \times 5 = 5 \times 7 = 5 \times 5 \times 5 = \ldots$. It's a little weird that $5$ is its own square and cube, but so it is.

As I was taught here a few weeks ago, this domain is Noetherian, which means that ideals can be factorized finitely and uniquely.

Then $\langle 5 \rangle$ is not a prime ideal but can be factorized as a product of prime ideals. One factor might be $\langle 3, 5, 7 \rangle$, but since $3 = -7$, then $\langle 3, 5 \rangle$ will do. And then $\langle 3, 5 \rangle \langle 3, 5 \rangle = \langle 9, 5 \rangle$, but $9 = -1$, which therefore means $\langle 5 \rangle = \langle 3, 5 \rangle^2$.

Is this right? Or did I go wrong somewhere?

Answer 1

Notherian isn't enough for unique factorization of ideals, you need a Dedekind domain for that, which $\Bbb Z_{10}$ is not. In this case, for example, $(5)$ is a prime ideal because $\Bbb Z_{10}/(5)\cong\Bbb Z_5$, and in fact you can get this from the basic ring isomorphism theorems:

$$\Bbb Z/5\cong \Bbb Z/10 \bigg/ (5\Bbb Z/10)$$

Since $(5)\supseteq (10)$ is a prime ideal of $\Bbb Z$.

Answer 2

Well, for one thing, you forgot that domains like $\textbf{Z}_{10}$ have numbers that are prime but not reducible. With the calculations you have already done, you can verify that whenever $ab = 5$, either $a = 5$ or $b = 5$. That means prime. Plus you have also observed what happens with $a = b = 5$.

Prime numbers generate prime ideals. And prime ideals are maximal ideals. When you deleted $7$ from $\langle 3, 5, 7 \rangle$, you realized you were removing a redundant generator. But you overlooked that $\langle 3 \rangle$ is the entire ring, and so $\langle 3, 5 \rangle$ is a principal ideal after all, and trivially so.

Of course this does not mean that you're insane, merely that rings such as these are kind of insane by the standards of $\textbf{Z}$ and quadratic integer rings.

Answer 3

A Noetherian ring is one with no infinite ascending chain of ideals. In $\mathbb{Z}_{10}$ there is only two possibilities;

• $(0) \subseteq (2)=2\mathbb{Z}_{10} \subseteq (1)=\mathbb{Z}_{10} $
• $(0) \subseteq (5)=5\mathbb{Z}_{10} \subseteq (1)=\mathbb{Z}_{10} $

In an Noetherian wring, every element can be factored into irreducibles but not necessarily unique. There's not enough room for things to get complicated.

• $\mathbb{Z}_{10}/5\mathbb{Z}_{10}= \mathbb{Z}_5$ which is a field, so that $(5)$ is prime.

Many interesting rings arise in commutative algebra where these definitions can be important

• $\mathbb{Q}[x,y,z]/(x^2 + y^2 + z^2 -1)$

• the span $\langle x^a y^b : b > \sqrt{2} a > 0\rangle $ is not finitely generated.