Let $k$ be an algebraically closed field, and let $C\subset\mathbb{A}_k^2$ be a curve (i.e. irreducible of dimension 1), given by some prime ideal $\mathfrak{p}\in\mathrm{Spec}\, k[x,y]$. Must $\mathfrak{p}$ be principal?
Are affine plane curves given by principal ideals?
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abstract-algebra
algebraic-geometry
commutative-algebra
2 Answers
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I agree the answer posted by Chirantan Chowdhury. In fact we have the following proposition arising from Krull's Hauptidealsatz.
A variety $Y$ in $\mathbb{A}^n$ has dimension $n-1$ iff it is the zero of some single nonconstant irreducible polynomial in $\mathcal{k}[x_1, \cdots, x_n]$.
See Hartshorne pp7. However if codim$Y >1$, the answer is NO! For example, the curve $C$ in $\mathbb{A}^3$ given parametrically by $x_1=t^3,x_2=t^4,x_3=t^5$. $C$ can NOT be generated by two elements. This induces a very important notion complete intersection in algebraic geometry.
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Yes i think it is as of dimension 1 , the height of the prime ideal is 1 and in a UFD ($Spec(k[x,y])$, the height one prime ideals are principal.