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Over a field of characteristic zero, an alternating polynomial changes sign when two variables are interchanged. It vanishes when two coordinates agree, so each factor $x_i-x_j$ divides it. These distinct linear factors are relatively prime in the polynomial ring, hence their product, the Vandermonde determinant, divides it. The quotient is a symmetric polynomial.

For a positive integer $N$, all binomial coefficients $\left(\genfrac{}{}{0ex}{}{N}{j}\right)$ with $0<j<N$ are even exactly when $N$ is a power of two. Write the binary expansion $N={\sum }_{i\in I}{2}^i$. In the polynomial ring ${\mathbb{F}}_2[t]$, repeated squaring givesIf $I$ has one element, there are no interior terms. If it has more than one, the coefficient of $t^{2^{\min I}}$ is one and that exponent lies strictly between zero and $N$. This characterization gives a useful parity obstruction directly from the binary expansion.

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

For an element $a$ of a commutative $R$-algebra, evaluation on a polynomial ring is the ring homomorphism

If a finite group has no nontrivial homomorphism to ${\mathbb{C}}^{\times }$, its polynomial invariant ring is a unique factorization domain. Factor an invariant in the ambient polynomial ring and collect its irreducible factors into orbit products. Each orbit product transforms by a linear character, hence is invariant. It is prime in the invariant ring, and the invariant factorization is a product of these primes.

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.

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 $-\infty$.

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 the polynomial ring $A=k[X_1,\dots ,X_r]$ in characteristic zero, antisymmetrization identifies $H{H}^{\ast }(A,A)$ with ${\bigwedge }_A^{\ast }{\mathrm{Der}}_k(A)$. The map on a wedge of $p$ derivations is $\frac{1}{p!}{\sum }_{\sigma \in S_p}\mathrm{sgn}(\sigma ){\prod }_j{D}_{\sigma (j)}(a_j)$. The Hochschild cup product becomes the exterior product, and the left Gerstenhaber bracket becomes the left Schouten-Nijenhuis bracket. This assertion requires a smoothness hypothesis when generalized beyond polynomial rings.

The polynomial ring $\mathbb{Z}[X]$ is a unique factorization domain by Gauss lemma for polynomials, but its ideal $(2,X)$ is not principal. Any generator would divide both 2 and $X$, hence be a unit: a divisor of 2 has degree zero, and a constant dividing $X$ is $\pm 1$. Yet evaluation at zero followed by reduction modulo 2 annihilates the ideal and not 1. It embeds in the principal ideal domain $\mathbb{Q}[X]$, illustrating that the property need not pass to a subring.

The Kähler differentials of a polynomial ring are free on the differentials of its variables. Assigning arbitrary values to these generators uniquely defines a derivation. Applying the Conormal exact sequence for Kähler differentials to a quotient gives its presentation by the formal derivatives of its defining equations.

The Koszul complex on $f_1,\dots ,f_r$ has term $K_i={\bigwedge }^i{R}^r$ and differential $\partial (e_{j_1}\wedge \cdots \wedge e_{j_i})={\sum }_a(-1{)}^{a-1}{f}_{j_a}{e}_{j_1}\wedge \cdots \wedge \widehat{e_{j_a}}\wedge \cdots \wedge e_{j_i}$. Its degree-zero homology is $R/(f_1,\dots ,f_r)$. For a regular sequence it is exact in positive degrees, hence a Koszul resolution of that quotient. The variables of a polynomial ring form a regular sequence, so this resolves the coefficient field in length equal to the number of variables.

For a regular sequence $f_1,\dots ,f_r$ in a commutative ring $R$, its Koszul complex is the finite free projective resolution of $R/(f_1,\dots ,f_r)$ with terms ${\bigwedge }^j{R}^r$ and boundary map contracting by $(f_1,\dots ,f_r)$. Exactness follows by induction: append $f_r$ using the mapping cone for multiplication by $f_r$, which is injective on the preceding quotient. For a polynomial ring viewed as a bimodule, use the regular sequence $X_{i,\ell }-X_{i,r}$ in the enveloping algebra.

The kernel of $x\mapsto t^3$, $y\mapsto t^4$, $z\mapsto t^5$ is generated by $y^2-xz$, $yz-x^3$, $z^2-x^{2}y$. Reducing these three relations leaves a unique normal form $A_0(x)+A_1(x)y+A_2(x)z$: its images have distinct exponents modulo three. The curve ring is therefore finite free of rank three over $k[x]$ and has dimension one. In a three-variable polynomial ring its ideal has height two, but its three independent quadratic initial forms force at least three generators. This is a concrete failure of generation by codimension many equations.

Every unital $k$-subalgebra $A\subseteq k[T,T^{-1}]$ is a Noetherian ring. If $A\subseteq k[T]$ or $k[T^{-1}]$, a nonconstant element $f\in A$ makes that ambient polynomial ring a finitely generated module over $k[f]$. Otherwise $A$ contains an $f$ with both positive and negative exponents, and both $T$ and $T^{-1}$ are integral elements over $k[f]$. In either case $A$ is a submodule of a finite module over the Noetherian ring $k[f]$, and every ideal of $A$ is finitely generated over $k[f]$, hence over $A$.