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For any commutative ring $A$ and a monic polynomial $F$ of positive degree $D$, monic polynomial division over a ring gives unique representatives of degree less than $D$. Thus the classes of $1,T,\dots ,T^{D-1}$ form a basis of a module over $A$. This is a finite free algebra, hence flat, and the map $A\to A[T]/(F)$ is injective. The monic hypothesis makes leading-degree arguments valid even when $A$ has zero divisors.

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.

A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expressionwith $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a basis of a module and $M$ is the free module on $S$.

If $S$ freely generates $M$ and $f:S\to N$ is any function, defineUnique coordinates make $\varphi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the universal property of a free module.

Now let $T$ generate the free module $M$ and $|T|=m$. Choose a maximal ideal $\mathfrak{m}$ of the nontrivial ring $R$. Then $k=R/\mathfrak{m}$ is a field, andis a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. ThereforeIn particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the rank of a free module $\mathrm{rk}M$.

A Euclidean domain is a principal ideal domain. By the submodule theorem for free modules over a principal ideal domain, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives

The primary decomposition theorem for finitely generated modules over a principal ideal domain states thatwhere $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.

Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb{Z}$ gives an invariant-factor decompositionIts exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whenceBut $d_r\le |H|$, and equality forces all earlier factors to be trivial. Thus

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.