$\mathbb{Z}_{(2)}$ has one maximal ideal

Question

My lecture notes state that the set $\mathbb{Z}_{(2)}$, defined as

$$\mathbb{Z}_{(2)}:=\left\{\frac{a}{b}\in\mathbb{Q}\mathrel{}\middle|\mathrel{}\gcd(a,b)=1\text{ and } 2\nmid b\right\}$$ has a unique maximal ideal generated by $\left<2\right>$. Is this example trying to say that the set of even integers is the maximal ideal, and if so, why is this maximal. Can anyone prove that this is indeed a maximal ideal. Also in the set $\mathbb{Z}$, both $2\mathbb{Z}$ and $3\mathbb{Z}$ are maximal ideals, but I thought a maximal ideal is an ideal which contains all other ideals.

Answer 1

You can show that of the complement of $2 \mathbf{Z}_{(2)}$ is $\mathbf{Z}_{(2)}^{\times}$, showing thereby that $2 \mathbf{Z}_{(2)}$ is the unique maximal ideal of $\mathbf{Z}_{(2)}$. This show that $\mathbf{Z}_{(2)}$ is a local ring.

"Alternatively", consider the "map" from $\mathbf{Z}_{(2)}$ to $\mathbf{F}_2$ defined by sending an element $x= \frac{a}{b}$ with $a,b$ integers and $2$ not dividing $b$ to $\overline{a} \overline{b}^{-1}$, where $\overline{x}$ is the image in $\mathbf{F}_2$ of the integer $x$. Show that this map is well defined, that is, that it does not depend on the writing $x= \frac{a}{b}$ with $a,b$ integers and $2$ not dividing $b$, and that it is a ring morphism. Show that its kernel is the ideal $2 \mathbf{Z}_{(2)}$. As this morphism is surjective, this implies passing to the quotient that the induced map $\mathbf{Z}_{(2)} / 2 \mathbf{Z}_{(2)} \to \mathbf{F}_2$ is an isomorphism. As $\mathbf{F}_2$ is a field, this shows that $2 \mathbf{Z}_{(2)}$ is a maximal ideal of $\mathbf{Z}_{(2)}$.

Intuition. More generally instead of $2$, let $p$ be a prime. What is $\mathbf{Z}_{(p)}$ ? The ring $\mathbf{Z}$ is a ring where only $\pm 1$ are invertible. We want to construct a bigger ring, containing $\mathbf{Z}$, and in which every guy from $\mathbf{Z}$ that is not divisible by $p$ will become invertible. This ring is $\mathbf{Z}_{(p)}$. (For any ring, $A^{\times}$ denotes the set of invertible elements of $A$.) Then, $\mathbf{Z}_{(p)}^{\times} = \mathbf{Z}_{(p)} \backslash p\mathbf{Z}_{(p)}$.

Remark. More generally, if $A$ is a commutative ring with unit and if $\mathfrak{p}$ is a prime ideal of $A$, with the same arguments, you have an isomorphism $A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} \to \textrm{Frac}(A/\mathfrak{p})$, where $A_{\mathfrak{p}}$ is the localisation of $A$ with respect to the multiplicative subset $S = A \backslash \mathfrak{p}$ of $A$. The ring $A_{\mathfrak{p}}$ is also called by abuse of language the location of $A$ at the prime ideal $\mathfrak{p}$, or also more simply, the localization of $A$ a $\mathfrak{p}$. This vocabulary applies of course to $A = \mathbf{Z}$ and $\mathfrak{p} = 2 \mathbf{Z}$.

Answer 2

The maximal ideal is $2\mathbb{Z}_{(2)}$.

Show it is maximal by showing that $\mathbb{Z}_{(2)}/2\mathbb{Z}_{(2)}$ is a field.

To show that it is the only maximal ideal, there are many ways :

• Show that $J(\mathbb{Z}_{(2)})=2\mathbb{Z}_{(2)}$ where $J$ is the Jacobson ideal.
• Use the following theorem : If $R$ is a commutative ring and $M$ is a maximal ideal of $R$, then the following are equivalent :

(1) $R$ has a unique maximal ideal (such a ring is called local)

(2) $R\backslash M=R^{\times}$

(3) $\forall a,b\not\in R^{\times},a+b\not\in R^{\times}$.

Answer 3

Hint $\ $ All odd primes are units so all nonunits are associates of $\,2^n,\,$ so all ideals can be generated by powers of $\,2,\,$ hence are principal, generated by their least power of $\,2;\,$ for example assuming $\,i\le j\le k\le \ldots\,$ then $\,(2^i,2^j,2^k,\ldots)\, =\, 2^i(1, 2^{j-i},2^{k-i},\ldots)\, =\, (2^i).$