The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.26 Examples of spectra of rings

In this section we put some examples of spectra.

Example 10.26.1. In this example we describe $X = \mathop{\mathrm{Spec}}(\mathbf{Z}[x]/(x^2 - 4))$. Let $\mathfrak {p}$ be an arbitrary prime in $X$. Let $\phi : \mathbf{Z} \to \mathbf{Z}[x]/(x^2 - 4)$ be the natural ring map. Then, $ \phi ^{-1}(\mathfrak p)$ is a prime in $\mathbf{Z}$. If $ \phi ^{-1}(\mathfrak p) = (2)$, then since $\mathfrak p$ contains $2$, it corresponds to a prime ideal in $\mathbf{Z}[x]/(x^2 - 4, 2) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x^2)$ via the map $ \mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, 2)$. Any prime in $(\mathbf{Z}/2\mathbf{Z})[x]/(x^2)$ corresponds to a prime in $(\mathbf{Z}/2\mathbf{Z})[x]$ containing $(x^2)$. Such primes will then contain $x$. Since $(\mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x)$ is a field, $(x)$ is a maximal ideal. Since any prime contains $(x)$ and $(x)$ is maximal, the ring contains only one prime $(x)$. Thus, in this case, $\mathfrak p = (2, x)$. Now, if $ \phi ^{-1}(\mathfrak p) = (q)$ for $q > 2$, then since $\mathfrak p$ contains $q$, it corresponds to a prime ideal in $\mathbf{Z}[x]/(x^2 - 4, q) \cong (\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)$ via the map $ \mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, q)$. Any prime in $(\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)$ corresponds to a prime in $(\mathbf{Z}/q\mathbf{Z})[x]$ containing $(x^2 - 4) = (x -2)(x + 2)$. Hence, these primes must contain either $x -2$ or $x + 2$. Since $(\mathbf{Z}/q\mathbf{Z})[x]$ is a PID, all nonzero primes are maximal, and so there are precisely 2 primes in $(\mathbf{Z}/q\mathbf{Z})[x]$ containing $(x-2)(x + 2)$, namely $(x-2)$ and $(x + 2)$. In conclusion, there exist two primes $(q, x-2)$ and $(q, x + 2)$ since $2 \neq -2 \in \mathbf{Z}/(q)$. Finally, we treat the case where $\phi ^{-1}(\mathfrak p) = (0)$. Notice that $\mathfrak p$ corresponds to a prime ideal in $\mathbf{Z}[x]$ that contains $(x^2 - 4) = (x -2)(x + 2)$. Hence, $\mathfrak p$ contains either $(x-2)$ or $(x + 2)$. Hence, $\mathfrak p$ corresponds to a prime in $\mathbf{Z}[x]/(x - 2)$ or one in $\mathbf{Z}[x]/(x + 2)$ that intersects $\mathbf{Z}$ only at $0$, by assumption. Since $\mathbf{Z}[x]/(x - 2) \cong \mathbf{Z}$ and $\mathbf{Z}[x]/(x + 2) \cong \mathbf{Z}$, this means that $\mathfrak p$ must correspond to $0$ in one of these rings. Thus, $\mathfrak p = (x - 2)$ or $\mathfrak p = (x + 2)$ in the original ring.

Example 10.26.2. In this example we describe $X = \mathop{\mathrm{Spec}}(\mathbf{Z}[x])$. Fix $\mathfrak p \in X$. Let $\phi : \mathbf{Z} \to \mathbf{Z}[x]$ and notice that $\phi ^{-1}(\mathfrak p) \in \mathop{\mathrm{Spec}}(\mathbf{Z})$. If $\phi ^{-1}(\mathfrak p) = (q)$ for $q$ a prime number $q > 0$, then $\mathfrak p$ corresponds to a prime in $(\mathbf{Z}/(q))[x]$, which must be generated by a polynomial that is irreducible in $(\mathbf{Z}/(q))[x]$. If we choose a representative of this polynomial with minimal degree, then it will also be irreducible in $\mathbf{Z}[x]$. Hence, in this case $\mathfrak p = (q, f_ q)$ where $f_ q$ is an irreducible polynomial in $\mathbf{Z}[x]$ that is irreducible when viewed in $(\mathbf{Z}/(q) [x])$. Now, assume that $\phi ^{-1}(\mathfrak p) = (0)$. In this case, $\mathfrak p$ must be generated by nonconstant polynomials which, since $\mathfrak p$ is prime, may be assumed to be irreducible in $\mathbf{Z}[x]$. By Gauss' lemma, these polynomials are also irreducible in $\mathbf{Q}[x]$. Since $\mathbf{Q}[x]$ is a Euclidean domain, if there are at least two distinct irreducibles $f, g$ generating $\mathfrak p$, then $1 = af + bg$ for $a, b \in \mathbf{Q}[x]$. Multiplying through by a common denominator, we see that $m = \bar{a}f + \bar{b} g$ for $\bar{a}, \bar{b} \in \mathbf{Z}[x]$ and nonzero $m \in \mathbf{Z}$. This is a contradiction. Hence, $\mathfrak p$ is generated by one irreducible polynomial in $\mathbf{Z}[x]$.

Example 10.26.3. In this example we describe $X = \mathop{\mathrm{Spec}}(k[x, y])$ when $k$ is an arbitrary field. Clearly $(0)$ is prime, and any principal ideal generated by an irreducible polynomial will also be a prime since $k[x, y]$ is a unique factorization domain. Now assume $\mathfrak p$ is an element of $X$ that is not principal. Since $k[x, y]$ is a Noetherian UFD, the prime ideal $\mathfrak p$ can be generated by a finite number of irreducible polynomials $(f_1, \ldots , f_ n)$. Now, I claim that if $f, g$ are irreducible polynomials in $k[x, y]$ that are not associates, then $(f, g) \cap k[x] \neq 0$. To do this, it is enough to show that $f$ and $g$ are relatively prime when viewed in $k(x)[y]$. In this case, $k(x)[y]$ is a Euclidean domain, so by applying the Euclidean algorithm and clearing denominators, we obtain $p = af + bg$ for $p, a, b \in k[x]$. Thus, assume this is not the case, that is, that some nonunit $h \in k(x)[y]$ divides both $f$ and $g$. Then, by Gauss's lemma, for some $a, b \in k(x)$ we have $ah | f$ and $bh | g$ for $ah, bh \in k[x]$. By irreducibility, $ah = f$ and $bh = g$ (since $h \notin k(x)$). So, back in $k(x)[y]$, $f, g $ are associates, as $\frac{a}{b} g = f$. Since $k(x)$ is the fraction field of $k[x]$, we can write $g = \frac{r}{s} f $ for elements $r , s \in k[x]$ sharing no common factors. This implies that $sg = rf$ in $k[x, y]$ and so $s$ must divide $f$ since $k[x, y]$ is a UFD. Hence, $s = 1$ or $s = f$. If $s = f$, then $r = g$, implying $f, g \in k[x]$ and thus must be units in $k(x)$ and relatively prime in $k(x)[y]$, contradicting our hypothesis. If $s = 1$, then $g = rf$, another contradiction. Thus, we must have $f, g$ relatively prime in $k(x)[y]$, a Euclidean domain. Thus, we have reduced to the case $\mathfrak p$ contains some irreducible polynomial $p \in k[x] \subset k[x, y]$. By the above, $\mathfrak p$ corresponds to a prime in the ring $k[x, y]/(p) = k(\alpha )[y]$, where $\alpha $ is an element algebraic over $k$ with minimum polynomial $p$. This is a PID, and so any prime ideal corresponds to $(0)$ or an irreducible polynomial in $k(\alpha )[y]$. Thus, $\mathfrak p$ is of the form $(p)$ or $(p, f)$ where $f$ is a polynomial in $k[x, y]$ that is irreducible in the quotient $k[x, y]/(p)$.

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 $R$.

Comments (2)

Comment #3410 by Kazuki Masugi on

Typo in the Example10.26.4. ; " must be a prime ideal …" should be " must be a prime ideal…".

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00EX. Beware of the difference between the letter 'O' and the digit '0'.