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It is easy to make counterexamples examples of such subrings. For example, take $A=k[x,y]$ and consider the subring

$$ B=k[x^a y^b : 0\le \frac{b}{a}<\sqrt{2}]. $$Geometrically, $B$ is spanned by monomials whose exponent vectors lie below the line $y=\sqrt{2}x$.

I think your question is more quite interesting in the setting where $B=A^G$ B=A^G\subset A$ is the invariant ring of some group action on $A$, A$ (or equivalently, on the space $X=\mbox{Spec }A$. A$). In many cases this subalgebra is finitely generated, which allows one can define a quotient space $X/G$ by $Y=\mbox{Spec }A^G$ with many good properties. This happens for example if $G$ is finite or reductive. However, as shown by Nagata's famous counterexample to Hilbert's 14th problem, $A^G$ may be infinitely generated, so the problem of defining such quotients in general is subtle. (Nagata's construction is indeed very geometrical, but a bit too complicated to restate here).
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Let

It is easy to make counterexamples of such subrings. For example, take $A=k[x,y]$ and consider the subring

$$ B=k[x^a y^b : 0\le \frac{b}{a}<\sqrt{2}]. $$Geometrically, $B$ is spanned by monomials whose exponent vectors lie below the line $y=\sqrt{2}x$. Since

I think your question is more interesting in the setting where $B=A^G$ is the invariant ring of some group action on $A$, or equivalently, on the space $X=\mbox{Spec }A$. In many cases this line does not have any lattice pointssubalgebra is finitely generated, which allows one can define a quotient space $B$ X/G$ by $Y=\mbox{Spec }A^G$ with many good properties. This happens for example if $G$ is finite or reductive. However, as shown by Nagata's famous counterexample to Hilbert's 14th problem, $A^G$ may be infinitely generated, so the problem of defining such quotients in general is subtle. (Nagata's construction is indeed very geometrical, but a bit too complicated to restate here).
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Let $A=k[x,y]$ and consider the subring

$$ B=k[x^a y^b : 0\le \frac{b}{a}<\sqrt{2}]. $$Geometrically, $B$ is spanned by monomials whose exponent vectors lie below the line $y=\sqrt{2}x$. Since this line does not have any lattice points, $B$ is infinitely generated.