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Suppose we have a ring $A$, and an open subset $U\subseteq\mathrm{Spec}\, A$ such that $(U,\mathcal{O}_{\mathrm{Spec}\,A}\vert_U)$ is affine. Must $U=D(f)$ for some $f\in A$? The contrary is true, by a well-known theorem in basic algebraic geometry.

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    See http://mathoverflow.net/questions/133470/affine-open-subset-of-affine-scheme – 2017-02-06
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    I would not say that «the contrary is *clearly* true», really: it is somewhat nontrivial to prove the correct statements and/or to give examples, and therefore the adverb clearly seems quite out of place. (Additionally, saying "the contrary is true" is most surely not what you meant to say, but something like "there exist affine open sets which are not of this form", as you surely do not think that *no* set is of that form!) – 2017-02-06
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    What I meant was that all distinguished opens must be affine – though I suppose the proof is actually nontrivial, it is clear in the sense that it is one of the first things people learn once the concept of an affine scheme is introduced. – 2017-02-06
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    Nothing can be clear and at the same time require a nontrivial proof — that's an oxymoron. You say that "it is clear in thatr it is one of the first things people learn" but it is not one of the first things people learn. What they do learn first is what a distinguished open set is, that they are a basis, and so on, but certainly they do not learn that all affine open sets are of that form —mostly, because they are not— and they certainly do not get told at that point how to check that they are not, because at that point they cannot do it. – 2017-02-06
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    Lots of frustration can be eliminated by stopping to say that things which are neither "clear" nor "obvious" nor "evident" are clear, obvious or evident. At the very least, one should never use those words to describe a fact if one oneself cannot prove it. – 2017-02-06
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    After your edit, the question makes no sense. You ask if $U$ must be of the form $D(f)$ only to immediately tell us that a theorem —and a well-known and basic one, to boot!— tells you that the contrary is true. (And what does that theorem say, that $U$ is *not* of that form? In that case, the theorem is false) – 2017-02-06
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    I think there is a very nice counterexample, I might give it as an answer, but is is a mixture of the answer here http://mathoverflow.net/questions/7153/open-affine-subscheme-of-affine-scheme-which-is-not-principal, not the accepted answer, the other one. Once you think about it the example he gives is pretty canonical. To verify some of the details see this answer http://math.stackexchange.com/questions/2120108/example-of-affine-scheme-x-operatornamespec-r-and-irreducible-u-subset-x/2120695#2120695 which verifies that a different set is not a localization. – 2017-02-06

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Here is a counter example, see my comments above about its origin.

Let $$A=k[x,y,u,v]/(x^2u+y^2v-xy)$$ Then in $A$ we have the relation $$x^2u+y^2v=xy$$

Let $$U=D(x)\cup D(y)$$

We must show

$$\text{1) $U$ is affine.}$$ $$\text{2) $U \neq D(f)$ }$$

Note that $A$ is an integral domain, as $x^2u+y^2v-xy$ being linear in $u$ and $v$ is irreducible. Thus the definition of localization is simplified, maps between localizations are injective and $A_f\subseteq A_g$ iff $g|f$.

Next note that in any sheaf $\mathcal{F}$ with opens $V$ and $W$, $\mathcal{F}(V\cup W)$ is the pullback of $\mathcal{F}(V)$ and $\mathcal{F}(W)$ over $\mathcal{F}(V\cap W)$. Which means, taking into account that maps between localizations are injective, that $$\mathcal{O}_X(U)=\mathcal{O}_X(D(x))\cap \mathcal{O}_X(D(y))=A_x\cap A_y$$

Now to see that $U$ is affine note that

$$\text{$D(x)=X_x=U_x$ is affine}$$

$$\text{$D(y)=X_y=U_y$ is affine}$$

and $$U=U_x\cup U_y.$$

Thus in order to apply the criteria for affineness we need only show that $x$ and $y$ generate the unit ideal in $A_x\cap A_y$.

This is a consequence of the equation

$$x\left(\frac{u}{y}\right)+y\left(\frac{v}{x}\right)=1$$

which is valid since

$$\frac{u}{y}=\frac{x-yv}{x^2}\in A_x\cap A_y$$

$$\frac{v}{x}=\frac{y-xu}{y^2}\in A_x\cap A_y$$ Thus $U$ is affine.

To see that $U\neq D(f)$ we must show that $$A_x\cap A_y\neq A_f$$

however $A_f\subseteq A_x\cap A_y$$ implies that

$$\text{$x=sf$ in $A$}$$ $$\text{$y=tf$ in $A$}$$

or $$\text{$x=sf+p(x^2u+y^2v-xy)$ in $k[x,y,u,v]$}$$ $$\text{$y=tf+q(x^2u+y^2v-xy)$ in $k[x,y,u,v]$}$$

which since the left had sides have degree $1$ gives $f\in k$, and thus $A_f=A$. However $A_x\cap A_y\neq A$ since the fraction $\frac{u}{y}$ is not in $A$.