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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

Introduce slack variables $s_1,s_2,s_3$. In the initial simplex algorithm dictionary, $x_2$ has the largest positive objective coefficient. The ratio test givesso $x_2$ enters and $s_3$ leaves. Solving the third constraint for $x_2$ givesThe objective becomesEvery reduced cost is now nonpositive, so the simplex optimality criterion givesThe other two slacks both equal one.

The dual linear program isThe vectoris dual feasible and has objective value $12$. By weak duality, it and the displayed primal point are optimal; they also satisfy complementary slackness.

Under the null hypothesis of independence between treatment and outcome, the expected counts in each treatment row are one half of the column totals:The Pearson chi-squared test of independence statistic isThe number of degrees of freedom isAt the 5% level the critical value is $7.81$. Since $8.30>7.81$, we reject the null hypothesis and find statistically significant evidence that the drug and placebo have different effects.

Gauss's law in electrostatics statesCylindrical symmetry and a coaxial Gaussian cylinder giveThe field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.

Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,The capacitance per unit length is thereforeThe electrostatic energy per unit length isSubstitution of $C$ verifies $U=Q^2/(2C)$.

Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A principal ideal domain is Noetherian, so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.

Conversely, supposeand $M/N$ is generated by the cosets ${\overline{m}}_1,\dots ,{\overline{m}}_s$. For any $m\in M$, its coset is an $R$-linear combination of the ${\overline{m}}_j$, soExpressing this remainder in terms of the $n_i$ shows thatThus $M$ is finitely generated if and only if both the submodule $N$ and the quotient module $M/N$ are finitely generated.