
Example 10.26.4. Consider the ring

$R = \{ f \in \mathbf{Q}[z]\text{ with }f(0) = f(1) \} .$

Consider the map

$\varphi : \mathbf{Q}[A, B] \to R$

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

$R_ a = \{ f \in \mathbf{Q}[z, \frac{1}{z-a}]\text{ with }f(0) = f(1) \} .$

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:

$R\subset R_ a\subset \mathbf{Q}[z, \frac{1}{z-a}], \quad R\subset \mathbf{Q}[z]\subset \mathbf{Q}[z, \frac{1}{z-a}].$

Recall (Lemma 10.16.5) 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

$\{ \mathfrak p \in \mathop{\mathrm{Spec}}(T) \mid S \cap \mathfrak p = \emptyset \} \subset \mathop{\mathrm{Spec}}(T).$

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

$\mathfrak {m}_0 = (z^2-z, z^3-z),$
$\mathfrak {m}_ a = ((z-1 + a)(z-a), (z^2-1 + a)(z-a)), \text{ and}$
$\mathfrak {m}_{1-a} = ((z-1 + a)(z-a), (z-1 + a)(z^2-a)).$

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

$(z^2 - z)^2(z - a)(\frac{a^2 - a}{z - a} + z) = (z^2 - z)^2(a^2 - a) + (z^2 - z)^2(z - a)z$

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 10.16.5, 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 of $a = 0$ 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 $R$.

Comment #741 by Wei Xu on

First,

The reason "$z^2-z = \frac{(z^2-z)(z-a)}{z-a}$" for the conclusion "so $z^2-z = \frac{(z^2-z)(z-a)}{z-a}$ is in $\mathfrak{m}_aR_a$" is not clear, because $\frac{1}{z-a}\notin R_a$. One fixed way is: $(z^2-z)(z-a)(\frac{a^2-a}{z-a}+z)=(z^2-z)(a^2-a)+(z^2-z)(z-a)z\in \mathfrak{m}_aR_a$, and since $(z^2-z)(z-a)z\in \mathfrak{m}_a\cap \mathfrak{m}_0\subseteq \mathfrak{m}_a$, we get $(z^2-z)\in \mathfrak{m}_aR_a$.

Second,

"the maximal ideal $(z-a)$" in

"while $\mathbf{Q}[z, \frac{1}{z-a}]$ is the localization of $\mathbf{Q}[z]$ at the maximal ideal $(z-a)$" should be "$z-a$."

Finally,

I do not quite understand the proof that the units of $R_a$ are $\mathbf{Q}^{\times}$:" If $\frac{f}{(z-a)^k}$ is a unit in $R_a$ ($f\in R$ and $k\geq0$ an integer), then we have for some $g\in R$ and some integer $\ell\geq0$. Since $R$ is an integral domain, this is equivalent to But $(z-a)^{k + \ell}$ is only an element of $R$ if $k = \ell = 0$; hence $f, g$ are units in $R$ as well."

Instead, since the set of units of $\mathbf{Q}[z,\frac{1}{z-a}]$ is $\{u(z-a)^k\mid u\in \mathbf{Q}^{\times},k\in \mathbb{Z}\}$, one can conclude that the set of units of $R_a$ is $\mathbf{Q}^{\times}$.

Comment #753 by on

OK, I tried to improve on this following your suggestions. Here is the commit. By the way, this example shows that trying to prove and explain things without using the language of schemes and the theory that goes with that is rather cumbersome.

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