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10.27 Examples of spectra of rings

In this section we put some examples of spectra.

Example 10.27.1. 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 10.27.2. 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 10.27.3. 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 10.27.4. 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 10.17.5) 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

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

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

\[ (z^2 - z)^2(z - a)\left(\frac{a^2 - a}{z - a} + z\right) = (z^2 - z)^2(a^2 - a) + (z^2 - z)^2(z - a)z \]

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 10.17.5, 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 $R$.

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…".

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