Localisation of Modules and Support Examples

By [L1] and [L2], $(\mathbb{Z}\mathrm{/}n\mathbb{Z}{)}_{(p)}\cong {\mathbb{Z}}_{(p)}{\otimes }_{\mathbb{Z}}\mathbb{Z}\mathrm{/}n\mathbb{Z}\cong {\mathbb{Z}}_{(p)}\mathrm{/}n{\mathbb{Z}}_{(p)}$.

If $a=0$, then $p\nmid n$, so $n\mathrm{/}1$ is a unit in ${\mathbb{Z}}_{(p)}$ by [L3]. Hence $n{\mathbb{Z}}_{(p)}={\mathbb{Z}}_{(p)}$, and step 1.1 gives $(\mathbb{Z}\mathrm{/}n\mathbb{Z}{)}_{(p)}=0$.

If $a>0$, write $n=p^{a}m$ with $(m,p)=1$. Then $m\mathrm{/}1$ is a unit in ${\mathbb{Z}}_{(p)}$, so $n{\mathbb{Z}}_{(p)}=p^a{\mathbb{Z}}_{(p)}$. Reduction modulo $p^a$ identifies ${\mathbb{Z}}_{(p)}\mathrm{/}{p}^a{\mathbb{Z}}_{(p)}$ with $\mathbb{Z}\mathrm{/}{p}^a\mathbb{Z}$, so step 1.1 gives $(\mathbb{Z}\mathrm{/}n\mathbb{Z}{)}_{(p)}\cong \mathbb{Z}\mathrm{/}{p}^a\mathbb{Z}$.

For a class $q+\mathbb{Z}\in \mathbb{Q}\mathrm{/}\mathbb{Z}$, if some integer prime to $p$ kills it then [L4] makes it zero in the localisation; this happens exactly for the torsion of order prime to $p$. On the other hand a class of order $p^r$ cannot be killed by any denominator outside $(p)$, so it survives. Therefore $(\mathbb{Q}\mathrm{/}\mathbb{Z}{)}_{(p)}$ is exactly the $p$-primary torsion subgroup.

By [L1], $[a]\in \ker (M\to S_2^{-1}M)$ exactly when $2^r[a]=0$ in $\mathbb{Z}\mathrm{/}12\mathbb{Z}$ for some $r\ge 0$. This happens for $[0],[3],[6],[9]$, and for no other class, because $12\mid 2^{r}a$ is possible exactly when the odd part of $a$ is divisible by $3$.

Likewise, $[a]\in \ker (M\to S_3^{-1}M)$ exactly when $3^r[a]=0$ for some $r\ge 0$. This happens for $[0],[4],[8]$, and for no other class, because $12\mid 3^{r}a$ is possible exactly when the $2$-primary part of $a$ is divisible by $4$.

Steps 1.1 and 1.2 give the two kernels, and [L2] interprets them as the elements killed by the respective localisation maps.

Apply [L1] with $R=\mathbb{Z}$ and $I=(12)$. A prime ideal of $\mathbb{Z}$ contains $(12)$ exactly when its prime generator divides $12$, so the only such prime ideals are $(2)$ and $(3)$.

Therefore ${\mathrm{Supp}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}12\mathbb{Z})=\{(2),(3)\}$, as claimed.

By [L1], ${\mathrm{Supp}}_{\mathbb{Z}}(M)={\bigcup }_p{\mathrm{Supp}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}p\mathbb{Z})$.

Each summand contributes exactly the singleton $\{(p)\}$ by [L2], so the union in step 1.1 is the set of all nonzero prime ideals of $\mathbb{Z}$.

Hence ${\mathrm{Supp}}_{\mathbb{Z}}(M)=\{(p):p\text{ prime}\}$.

By [L3], every class in $\mathbb{Z}\mathrm{/}3\mathbb{Z}$ is $2$ times another class, so $IM=M$.

The module $M$ is nonzero because $[1{]}_3\mathrm{\ne }[0{]}_3$. Moreover $2\mathrm{\notin }J(\mathbb{Z})$ because $2\mathrm{\notin }(3)$, while $(3)$ is a maximal ideal of $\mathbb{Z}$ and [L1] makes $J(\mathbb{Z})$ lie in every maximal ideal.

Thus $IM=M$ and $M\mathrm{\ne }0$ hold with $I⊈J(R)$, exactly exhibiting why the Jacobson-radical hypothesis cannot be dropped.

The class of $1\mathrm{/}p$ is nonzero in $\mathbb{Q}\mathrm{/}{\mathbb{Z}}_{(p)}$, because $1\mathrm{/}p\mathrm{\notin }{\mathbb{Z}}_{(p)}$. Hence $M\mathrm{\ne }0$.

Every element of $M$ has the form $q+{\mathbb{Z}}_{(p)}$ with $q\in \mathbb{Q}$. Then $p(q\mathrm{/}p+{\mathbb{Z}}_{(p)})=q+{\mathbb{Z}}_{(p)}$, so multiplication by $p$ is surjective and therefore $pM=M$.

The module $M$ is not finitely generated. If classes $q_1+{\mathbb{Z}}_{(p)},\dots ,q_r+{\mathbb{Z}}_{(p)}$ generated $M$, choose $N$ so that every $q_i$ has denominator dividing $p^N$ modulo ${\mathbb{Z}}_{(p)}$. Then every generated class would also have denominator dividing $p^N$, but $1\mathrm{/}{p}^{N+1}+{\mathbb{Z}}_{(p)}$ would not lie in that span.

So $pM=M$ and $M\mathrm{\ne }0$ hold for a module that is not finitely generated, exactly showing why the finite-generation hypothesis is essential.

In $M\mathrm{/}\mathfrak{m}M=\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, the classes of $x$ and $y$ are nonzero and linearly independent over the residue field $k$, because every element of ${\mathfrak{m}}^2$ has total degree at least $2$.

Those same classes span $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, since every element of $\mathfrak{m}$ has image given by its linear part $ax+by$ modulo ${\mathfrak{m}}^2$.

By definition, $\mathfrak{m}=(x,y)R$, so $x$ and $y$ generate $M$. If a single element generated $M$, then its image would generate $M\mathrm{/}\mathfrak{m}M$, contradicting steps 1.1 and 1.2. Hence $\{x,y\}$ is a minimal generating set of $\mathfrak{m}$.

At the prime $(2)$, the localisation of $\mathbb{Z}\mathrm{/}6\mathbb{Z}$ is $\mathbb{Z}\mathrm{/}2\mathbb{Z}$, while the localisation of $\mathbb{Z}\mathrm{/}2\mathbb{Z}\oplus \mathbb{Z}\mathrm{/}3\mathbb{Z}$ is $\mathbb{Z}\mathrm{/}2\mathbb{Z}\oplus 0$, because $3$ becomes a unit. The localised map is therefore the identity on the surviving $\mathbb{Z}\mathrm{/}2\mathbb{Z}$ summand.

At the prime $(3)$, the same computation gives the identity $\mathbb{Z}\mathrm{/}3\mathbb{Z}\to 0\oplus \mathbb{Z}\mathrm{/}3\mathbb{Z}$.

At every other prime $(q)$, both $\mathbb{Z}\mathrm{/}6\mathbb{Z}$ and $\mathbb{Z}\mathrm{/}2\mathbb{Z}\oplus \mathbb{Z}\mathrm{/}3\mathbb{Z}$ localise to $0$ because both $2$ and $3$ become units.

Steps 1.1, 1.2, and 1.3 show that every prime localisation of $\varphi$ is an isomorphism, so [L1] gives that $\varphi$ is an isomorphism.

The module $M$ is free on countably many generators, so it is not finitely generated and therefore not finitely presented.

By [L2] and [L3], there is a $\mathbb{Z}[1\mathrm{/}p]$-linear map $\phi :S^{-1}M\to \mathbb{Z}[1\mathrm{/}p]$ with $\phi (e_n)=1\mathrm{/}{p}^n$ for every $n\ge 0$.

Suppose $\phi$ were in the image of the localisation-of-Hom map. Then there would be a homomorphism $f:M\to \mathbb{Z}$ and an integer $r\ge 0$ such that $\phi (e_n)=f(e_n)\mathrm{/}{p}^r$ for every $n$. Taking $n>r$ gives $f(e_n)=p^{r-n}$, impossible in $\mathbb{Z}$. Therefore $\phi$ is not in the image.

So the localisation-of-Hom map fails to be surjective for this non-finitely-presented source, exactly as warned by [L1].

The family $y=(1,1\mathrm{/}p,1\mathrm{/}{p}^2,\dots )$ is an element of ${\prod }_{n\ge 0}\mathbb{Z}[1\mathrm{/}p]$.

Suppose $y$ came from an element of $S^{-1}P$. By [L2], it would have the form $(a_0,a_1,\dots )\mathrm{/}{p}^r$ for some fixed $r\ge 0$ and integers $a_n$. Then the $n$th coordinate equation $a_n\mathrm{/}{p}^r=1\mathrm{/}{p}^n$ would force $a_n=p^{r-n}$ in $\mathbb{Z}$ for every $n>r$, impossible.

Therefore the displayed map is not surjective, so localisation does not commute with this infinite product. The contrast with [L1] is exactly the point of the example.