The Stacks project

Example 10.27.4. Consider the ring

Consider the map

defined by $\varphi (A) = z^2-z$ and $\varphi (B) = z^3-z^2$. It is easily checked that $(A^3 - B^2 + AB) \subset \mathop{\mathrm{Ker}}(\varphi )$ and that $A^3 - B^2 + AB$ is irreducible. Assume that $\varphi $ is surjective; then since $R$ is an integral domain (it is a subring of an integral domain), $\mathop{\mathrm{Ker}}(\varphi )$ must be a prime ideal of $\mathbf{Q}[A, B]$. The prime ideals which contain $(A^3-B^2 + AB)$ are $(A^3-B^2 + AB)$ itself and any maximal ideal $(f, g)$ with $f, g\in \mathbf{Q}[A, B]$ such that $f$ is irreducible mod $g$. But $R$ is not a field, so the kernel must be $(A^3-B^2 + AB)$; hence $\varphi $ gives an isomorphism $R \to \mathbf{Q}[A, B]/(A^3-B^2 + AB)$.

To see that $\varphi $ is surjective, we must express any $f\in R$ as a $\mathbf{Q}$-coefficient polynomial in $A(z) = z^2-z$ and $B(z) = z^3-z^2$. Note the relation $zA(z) = B(z)$. Let $a = f(0) = f(1)$. Then $z(z-1)$ must divide $f(z)-a$, so we can write $f(z) = z(z-1)g(z)+a = A(z)g(z)+a$. If $\deg (g) < 2$, then $h(z) = c_1z + c_0$ and $f(z) = A(z)(c_1z + c_0)+a = c_1B(z)+c_0A(z)+a$, so we are done. If $\deg (g)\geq 2$, then by the polynomial division algorithm, we can write $g(z) = A(z)h(z)+b_1z + b_0$ ($\deg (h)\leq \deg (g)-2$), so $f(z) = A(z)^2h(z)+b_1B(z)+b_0A(z)$. Applying division to $h(z)$ and iterating, we obtain an expression for $f(z)$ as a polynomial in $A(z)$ and $B(z)$; hence $\varphi $ is surjective.

Now let $a \in \mathbf{Q}$, $a \neq 0, \frac{1}{2}, 1$ and consider

This is a finitely generated $\mathbf{Q}$-algebra as well: it is easy to check that the functions $z^2-z$, $z^3-z$, and $\frac{a^2-a}{z-a}+z$ generate $R_ a$ as an $\mathbf{Q}$-algebra. We have the following inclusions:

Recall (Lemma 10.17.5) that for a ring T and a multiplicative subset $S\subset T$, the ring map $T \to S^{-1}T$ induces a map on spectra $\mathop{\mathrm{Spec}}(S^{-1}T) \to \mathop{\mathrm{Spec}}(T)$ which is a homeomorphism onto the subset

When $S = \{ 1, f, f^2, \ldots \} $ for some $f\in T$, this is the open set $D(f)\subset T$. We now verify a corresponding property for the ring map $R \to R_ a$: we will show that the map $\theta : \mathop{\mathrm{Spec}}(R_ a) \to \mathop{\mathrm{Spec}}(R)$ induced by inclusion $R\subset R_ a$ is a homeomorphism onto an open subset of $\mathop{\mathrm{Spec}}(R)$ by verifying that $\theta $ is an injective local homeomorphism. We do so with respect to an open cover of $\mathop{\mathrm{Spec}}(R_ a)$ by two distinguished opens, as we now describe. For any $r\in \mathbf{Q}$, let $\text{ev}_ r : R \to \mathbf{Q}$ be the homomorphism given by evaluation at $r$. Note that for $r = 0$ and $r = 1-a$, this can be extended to a homomorphism $\text{ev}_ r' : R_ a \to \mathbf{Q}$ (the latter because $\frac{1}{z-a}$ is well-defined at $z = 1-a$, since $a\neq \frac{1}{2}$). However, $\text{ev}_ a$ does not extend to $R_ a$. Write $\mathfrak {m}_ r = \mathop{\mathrm{Ker}}(\text{ev}_ r)$. We have

To verify this, note that the right-hand sides are clearly contained in the left-hand sides. Then check that the right-hand sides are maximal ideals by writing the generators in terms of $A$ and $B$, and viewing $R$ as $\mathbf{Q}[A, B]/(A^3-B^2 + AB)$. Note that $\mathfrak {m}_ a$ is not in the image of $\theta $: we have

The left hand side is in $\mathfrak m_ a R_ a$ because $(z^2 - z)(z - a)$ is in $\mathfrak m_ a$ and because $(z^2 - z)(\frac{a^2 - a}{z - a} + z)$ is in $R_ a$. Similarly the element $(z^2 - z)^2(z - a)z$ is in $\mathfrak m_ a R_ a$ because $(z^2 - z)$ is in $R_ a$ and $(z^2 - z)(z - a)$ is in $\mathfrak m_ a$. As $a \not\in \{ 0, 1\} $ we conclude that $(z^2 - z)^2 \in \mathfrak m_ a R_ a$. Hence no ideal $I$ of $R_ a$ can satisfy $I \cap R = \mathfrak m_ a$, as such an $I$ would have to contain $(z^2 - z)^2$, which is in $R$ but not in $\mathfrak m_ a$. The distinguished open set $D((z-1 + a)(z-a))\subset \mathop{\mathrm{Spec}}(R)$ is equal to the complement of the closed set $\{ \mathfrak {m}_ a, \mathfrak {m}_{1-a}\} $. Then check that $R_{(z-1 + a)(z-a)} = (R_ a)_{(z-1 + a)(z-a)}$; calling this localized ring $R'$, then, it follows that the map $R \to R'$ factors as $R \to R_ a \to R'$. By Lemma 10.17.5, then, these maps express $\mathop{\mathrm{Spec}}(R') \subset \mathop{\mathrm{Spec}}(R_ a)$ and $\mathop{\mathrm{Spec}}(R') \subset \mathop{\mathrm{Spec}}(R)$ as open subsets; hence $\theta : \mathop{\mathrm{Spec}}(R_ a) \to \mathop{\mathrm{Spec}}(R)$, when restricted to $D((z-1 + a)(z-a))$, is a homeomorphism onto an open subset. Similarly, $\theta $ restricted to $D((z^2 + z + 2a-2)(z-a)) \subset \mathop{\mathrm{Spec}}(R_ a)$ is a homeomorphism onto the open subset $D((z^2 + z + 2a-2)(z-a)) \subset \mathop{\mathrm{Spec}}(R)$. Depending on whether $z^2 + z + 2a-2$ is irreducible or not over $\mathbf{Q}$, this former distinguished open set has complement equal to one or two closed points along with the closed point $\mathfrak {m}_ a$. Furthermore, the ideal in $R_ a$ generated by the elements $(z^2 + z + 2a-a)(z-a)$ and $(z-1 + a)(z-a)$ is all of $R_ a$, so these two distinguished open sets cover $\mathop{\mathrm{Spec}}(R_ a)$. Hence in order to show that $\theta $ is a homeomorphism onto $\mathop{\mathrm{Spec}}(R)-\{ \mathfrak {m}_ a\} $, it suffices to show that these one or two points can never equal $\mathfrak {m}_{1-a}$. And this is indeed the case, since $1-a$ is a root of $z^2 + z + 2a-2$ if and only if $a = 0$ or $a = 1$, both of which do not occur.

Despite this homeomorphism which mimics the behavior of a localization at an element of $R$, while $\mathbf{Q}[z, \frac{1}{z-a}]$ is the localization of $\mathbf{Q}[z]$ at the maximal ideal $(z-a)$, the ring $R_ a$ is not a localization of $R$: Any localization $S^{-1}R$ results in more units than the original ring $R$. The units of $R$ are $\mathbf{Q}^\times $, the units of $\mathbf{Q}$. In fact, it is easy to see that the units of $R_ a$ are $\mathbf{Q}^*$. Namely, the units of $\mathbf{Q}[z, \frac{1}{z - a}]$ are $c (z - a)^ n$ for $c \in \mathbf{Q}^*$ and $n \in \mathbf{Z}$ and it is clear that these are in $R_ a$ only if $n = 0$. Hence $R_ a$ has no more units than $R$ does, and thus cannot be a localization of $R$.

We used the fact that $a\neq 0, 1$ to ensure that $\frac{1}{z-a}$ makes sense at $z = 0, 1$. We used the fact that $a\neq 1/2$ in a few places: (1) In order to be able to talk about the kernel of $\text{ev}_{1-a}$ on $R_ a$, which ensures that $\mathfrak {m}_{1-a}$ is a point of $R_ a$ (i.e., that $R_ a$ is missing just one point of $R$). (2) At the end in order to conclude that $(z-a)^{k + \ell }$ can only be in $R$ for $k = \ell = 0$; indeed, if $a = 1/2$, then this is in $R$ as long as $k + \ell $ is even. Hence there would indeed be more units in $R_ a$ than in $R$, and $R_ a$ could possibly be a localization of $R$.