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For a nonconstant polynomial over any field, use triangular substitutions $x_i=y_i+y_n^{B^i}$, $x_n=y_n$, where $B$ exceeds every exponent in its support. Base-$B$ weights distinguish all original monomials, so the transformed polynomial has a unique highest $y_n$ power with coefficient in $k^{\times }$. Rescale to make it monic. The monic polynomial quotient is finite free over $k[y_1,\dots ,y_{n-1}]$, yielding a finite flat projection. The nonlinear substitution works over finite fields as well as infinite ones; a generic linear-direction argument alone would not do so.

For an explicit basis extension in ${\mathbb{R}}^2$, start with $A=\{u\}$, where $u=(1,1)$. Its linear span is the line $y=x$. Adding $w=(1,-1)$ preserves linear independence, because $w$ is outside that line, and gives a basis. Every vector has the finite linear combinationThe vector $2u=(2,2)$ cannot be added while preserving linear independence: it is already in the linear span. Many choices outside the line work, so maximality does not select a unique basis.

For a finite-dimensional vector space with a supplied finite spanning set, a greedy algorithm scans the list, retaining each vector precisely when it is outside the linear span of those already retained. Starting with a linearly independent $A$, the same scan adjoins only the needed vectors. Retained vectors remain linearly independent, and every skipped vector is in the final linear span, so the result is a basis containing $A$. Finite reasoning needs no axiom of choice.

For an arbitrary vector space, this finite stopping argument is unavailable. Accepting Zorn's lemma asserts existence of a maximal element once every chain in a partial order has an upper bound; it supplies no general computational recipe. Over ZF, Zorn's lemma, the axiom of choice and the well-ordering theorem are equivalent. Particular infinite vector spaces can still have explicit bases, such as $1,x,x^2,\dots$ for the vector space of univariate polynomials over a specified field; infinity alone does not force a nonconstructive argument.^[ref]

Multiplication by a nonzero element of an integral domain is injective. If the ring is finite, that map is surjective, so its image contains one and the multiplier has a multiplicative inverse. Hence every nonzero element is invertible and the ring is a field.

A commutative Artinian ring has finite composition length as a module over itself. It has finitely many maximal ideals, and its Jacobson radical $J$ is nilpotent. The Chinese remainder theorem makes $R/J$ a finite product of fields; each Artinian module $J^i/J^{i+1}$ has finite-dimensional components over those fields. Adding their lengths along the finite radical filtration proves the assertion, and in particular the ascending chain condition.

Every finite subgroup $H$ of the multiplicative group of a field is cyclic. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the Lagrange root bound over a field gives $|H|\le m$. Since $m\le |H|$, equality holds and that element generates $H$.

A quiver has finite representation type over a field when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

A unital commutative ring generated by finitely many elements under addition and multiplication, starting from the image of the integers. Equivalently it is a quotient of a finite-variable polynomial ring over $\mathbb{Z}$. localization of a ring at one element remains finite type, since $R[1/r]\cong R[T]/(rT-1)$. Generation here is as an algebra, not merely as a field extension when the ring happens to be a field.

A field is formally real when $-1$ is not a finite sum of squares. The Artin-Schreier ordering criterion says this is equivalent to the existence of an ordering compatible with the field operations. Such a field has characteristic zero and embeds into a real closed field.

For a vector space $M$ over a field $k$, its formal power series module is $M[[t]]={\underleftarrow{\lim }}_{n}M{\otimes }_{k}k[t]/(t^n)$. It is complete for the filtration by $t^{n}M[[t]]$. The ordinary tensor product $M{\otimes }_{k}k[[t]]$ embeds in it and consists of series whose coefficients span a finite-dimensional vector subspace; it equals $M[[t]]$ if $M$ is finite-dimensional. The inclusion is proper for infinite-dimensional $M$, as shown by a series with linearly independent coefficients.

A finite-dimensional unital associative algebra over a field admitting a linear functional $\epsilon$ whose pairing $(a,b)\mapsto \epsilon (ab)$ is nondegenerate. The dual numbers with functional extracting the nilpotent coefficient are an example; semisimplicity is unnecessary.

In characteristic $p>0$, the displayed map on the cocenter of a finite-dimensional algebra is additive and Frobenius-semilinear. Expand a $p$th power into words: nonconstant cyclic-word orbits have length $p$ and cancel modulo commutators. Also $(ab{)}^p$ and $(ba{)}^p$ have the same cocenter class, so the map kills additive commutators and is well defined.

Over a perfect field with split semisimple quotient, the stable image of this map has dimension equal to the number of simple modules of the algebra. Indeed, the kernel of $C(A)\to C(A/J(A))$ has representatives in the nilpotent Jacobson radical, so an iterate kills it. On the semisimple algebra quotient, each matrix-block coordinate is a field trace, and $\mathrm{tr}(a^p)=\mathrm{tr}(a{)}^p$. For a split semisimple quotient, Frobenius is bijective on every coordinate. In a group algebra over an algebraically closed field, the stable image has the p-regular element conjugacy classes as its basis, proving the simple-module counting theorem.

A Galois character is a continuous one-dimensional Galois representation: the absolute Galois group acts by multiplication by a scalar in a coefficient field $F$. A quadratic character over a coefficient field of characteristic different from two has values in $\{1,-1\}$; after identifying this group with $\mathbb{Z}/2\mathbb{Z}$, its kernel fixes a quadratic extension, or the base field for the trivial character. Its restriction to an inertia group is trivial exactly when that extension is unramified at the corresponding place.

For a nonzero finitely generated algebra over a field, choose a finite-dimensional generating vector subspace $V$ containing $1$. The dimension is $\mathrm{GKdim}A={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_k{V}^n)/\log n$. Changing $V$ only rescales the filtration index by bounded factors. A nonzero finite-dimensional algebra has dimension zero; a polynomial ring in $r$ variables has dimension $r$. The zero algebra is often assigned dimension $-\infty$.

The global dimension of a ring is the supremum of the projective dimensions of all its modules, allowing infinity. It measures the longest projective resolution needed by any module. For a field it is zero, since every module is a vector space and hence free.

A global regular function is a global section of the structure sheaf of a scheme, defined on the entire scheme. These functions form its ring $H^0(X,{\mathcal{O}}_X)$. On an affine scheme this recovers the defining ring; on a positive-dimensional projective complete intersection it consists only of constants over the base field.

The only regular functions defined on all of a nonempty projective space over a field $k$ are the constants:For ${\mathbb{P}}^1$, this also follows by gluing $k[t]$ and $k[t^{-1}]$ inside $k[t,t^{-1}]$: their intersection is $k$.

A graded module over a graded ring $R={⨁}_i{R}_i$ is an $R$-module with a direct sum decomposition $M={⨁}_d{M}_d$ satisfying $R_i{M}_d\subseteq M_{i+d}$. A graded homomorphism preserves these degrees. Over a polynomial ring with positive variable degrees, a finitely generated module of this kind is bounded below and has finite-dimensional graded components over the coefficient field.

For a positively graded ring with degree-zero part a field, put $R_{+}={⨁}_{d>0}{R}_d$. A bounded-below graded module $M$ with $M=R_{+}M$ is zero: a nonzero homogeneous element of least degree would have to be a sum of positive-degree coefficients times elements of lower degree. Hence homogeneous lifts of a basis of $M/R_{+}M$ generate $M$.

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.