Distinguished open set isomorphic to affine variety in…
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Distinguished open set isomorphic to affine variety in higher-dimensional space
Recently, I asked about the usual proof that the ring of regular functions on distinguished open set D(f) on variety X \\subset \\mathbb{A}\^n is A(X)\_f where A(X) is the coordinate ring of X. In several places, including Hartshorne, there's a statement that looks highly related but is different: that D(f) is isomorphic to an affine variety X' \\subset \\mathbb{A}\^{n+1}, defined by \\{(x,t): (x,t)\\in A\^{n+1}, x \\in X, tf(x)-1 = 0\\}, and the coordinate ring of X' is A(X)\_f.
I have two questions:
1. Does this constitute another proof that the ring of regular functions on D(f) is A(X)\_f? It seems like it "obviously" should be, although I'm not sure what you need to do to formally show it.
2. This feels geometrically very unintuitive to me, though the algebra seems reasonably well motivated. How does one "see" that a distinguished open set isomorphic to an affine variety embedded in a space of one higher dimension?
#science
Recently, I asked about the usual proof that the ring of regular functions on distinguished open set D(f) on variety X \\subset \\mathbb{A}\^n is A(X)\_f where A(X) is the coordinate ring of X. In several places, including Hartshorne, there's a statement that looks highly related but is different: that D(f) is isomorphic to an affine variety X' \\subset \\mathbb{A}\^{n+1}, defined by \\{(x,t): (x,t)\\in A\^{n+1}, x \\in X, tf(x)-1 = 0\\}, and the coordinate ring of X' is A(X)\_f.
I have two questions:
1. Does this constitute another proof that the ring of regular functions on D(f) is A(X)\_f? It seems like it "obviously" should be, although I'm not sure what you need to do to formally show it.
2. This feels geometrically very unintuitive to me, though the algebra seems reasonably well motivated. How does one "see" that a distinguished open set isomorphic to an affine variety embedded in a space of one higher dimension?
#science
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