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Let $A=\mathbb{C}[x,y,z]$. This polynomial ring is a unique factorization domain of Krull dimension three. For a height-one prime in a unique factorization domain $\mathfrak{p}$, choose a nonzero element of it and an irreducible factor $q$ belonging to $\mathfrak{p}$. In a unique factorization domain, $q$ is prime, so $(q)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Strict containment would give a chain $(0)⊊(q)⊊\mathfrak{p}$ of length two. ThusIf $\mathfrak{p}$ has height three, it is maximal: any strictly larger proper prime would extend a length-three chain and contradict $\dim A=3$. The Weak Hilbert Nullstellensatz over the algebraically closed field $\mathbb{C}$ then givesfor some $a,b,c\in \mathbb{C}$. This proves both requested generator assertions.

For the monomial curve with exponents three, four and five, putAll three map to zero under the given parametrization, so $J\subseteq I=\ker \phi$. Reduce any monomial using $y^2\mapsto xz$, $yz\mapsto x^3$ and $z^2\mapsto x^{2}y$. Each reduction decreases the sum of the exponents of $y,z$, so it terminates. Every polynomial has, modulo $J$, the formIts image is $A_0(t^3)+t^4{A}_1(t^3)+t^5{A}_2(t^3)$. These three terms occupy different exponent classes modulo three, so they cannot cancel. A zero image forces each $A_i=0$. ConsequentlyThe quotient is the domain $\mathbb{C}[t^3,t^4,t^5]$, hence $I$ is a prime ideal. It is finite free over $\mathbb{C}[t^3]$, with basis of a module $1,t^4,t^5$, by the same normal-form argument. Its dimension is therefore one, using that integral extensions preserve Krull dimension. The polynomial-ring height and dimension formula givesFor the generator obstruction, let $\mathfrak{m}=(x,y,z)$. Since $I\subseteq {\mathfrak{m}}^2$, we have $\mathfrak{m}I\subseteq {\mathfrak{m}}^3$. The degree-two initial forms of $g_1,g_2,g_3$ areThey are linearly independent over $\mathbb{C}$ in ${\mathfrak{m}}^2/{\mathfrak{m}}^3$. Thus no nonzero constant linear combination of the $g_i$ belongs to $\mathfrak{m}I$, and their classes are independent in $I/\mathfrak{m}I$. Since the three generators span that quotient,Two global generators would span it by two vectors, which is impossible. This is the initial-form lower bound for ideal generators; it shows why height two does not force two generators, even though height one and height three behave as above.

The polynomial ring $\mathbb{C}[X,Y]$ is a two-dimensional unique factorization domain. Let $\mathfrak{p}$ be a prime ideal containing $(f,g)$. Since $f\ne 0$, the prime $\mathfrak{p}$ is nonzero. If it were not maximal, it would have height one, so the height-one prime in a unique factorization domain would give $\mathfrak{p}=(h)$ for an irreducible polynomial $h$. Then $h$ would divide both $f$ and $g$, contrary to the hypothesis. Every prime ofis therefore maximal.

The Hilbert basis theorem makes $R$ Noetherian, and it has Krull dimension zero by the preceding paragraph. The Noetherian dimension-zero criterion for an Artinian ring now shows that