## 111.24 Going up and going down

Definition 111.24.1. Let $\phi : A \to B$ be a homomorphism of rings. We say that the *going-up theorem* holds for $\phi $ if the following condition is satisfied:

for any ${\mathfrak p}, {\mathfrak p}' \in \mathop{\mathrm{Spec}}(A)$ such that ${\mathfrak p} \subset {\mathfrak p}'$, and for any $P \in \mathop{\mathrm{Spec}}(B)$ lying over ${\mathfrak p}$, there exists $P'\in \mathop{\mathrm{Spec}}(B)$ lying over ${\mathfrak p}'$ such that $P \subset P'$.

Similarly, we say that the *going-down theorem* holds for $\phi $ if the following condition is satisfied:

for any ${\mathfrak p}, {\mathfrak p}' \in \mathop{\mathrm{Spec}}(A)$ such that ${\mathfrak p} \subset {\mathfrak p}'$, and for any $P' \in \mathop{\mathrm{Spec}}(B)$ lying over ${\mathfrak p}'$, there exists $P\in \mathop{\mathrm{Spec}}(B)$ lying over ${\mathfrak p}$ such that $P \subset P'$.

Exercise 111.24.2. In each of the following cases determine whether (GU), (GD) holds, and explain why. (Use any Prop/Thm/Lemma you can find, but check the hypotheses in each case.)

$k$ is a field, $A = k$, $B = k[x]$.

$k$ is a field, $A = k[x]$, $B = k[x, y]$.

$A = {\mathbf Z}$, $B = {\mathbf Z}[1/11]$.

$k$ is an algebraically closed field, $A = k[x, y]$, $B = k[x, y, z]/(x^2-y, z^2-x)$.

$A = {\mathbf Z}$, $B = {\mathbf Z}[i, 1/(2 + i)]$.

$A = {\mathbf Z}$, $B = {\mathbf Z}[i, 1/(14 + 7i)]$.

$k$ is an algebraically closed field, $A = k[x]$, $B = k[x, y, 1/(xy-1)]/(y^2-y)$.

Exercise 111.24.3. Let $A$ be a ring. Let $B = A[x]$ be the polynomial algebra in one variable over $A$. Let $f = a_0 + a_1 x + \ldots + a_ r x^ r \in B = A[x]$. Prove carefully that the image of $D(f)$ in $\mathop{\mathrm{Spec}}(A)$ is equal to $D(a_0) \cup \ldots \cup D(a_ r)$.

Exercise 111.24.4. Let $k$ be an algebraically closed field. Compute the image in $\mathop{\mathrm{Spec}}(k[x, y])$ of the following maps:

$\mathop{\mathrm{Spec}}(k[x, yx^{-1}]) \to \mathop{\mathrm{Spec}}(k[x, y])$, where $k[x, y] \subset k[x, yx^{-1}] \subset k[x, y, x^{-1}]$. (Hint: To avoid confusion, give the element $yx^{-1}$ another name.)

$\mathop{\mathrm{Spec}}(k[x, y, a, b]/(ax-by-1))\to \mathop{\mathrm{Spec}}(k[x, y])$.

$\mathop{\mathrm{Spec}}(k[t, 1/(t-1)]) \to \mathop{\mathrm{Spec}}(k[x, y])$, induced by $x \mapsto t^2$, and $y \mapsto t^3$.

$k = {\mathbf C}$ (complex numbers), $\mathop{\mathrm{Spec}}(k[s, t]/(s^3 + t^3-1)) \to \mathop{\mathrm{Spec}}(k[x, y])$, where $x\mapsto s^2$, $y \mapsto t^2$.

## Comments (0)