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.
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