Polynomial ring

The polynomial ring $R[X]$ consists of polynomials in an indeterminate $X$ with coefficients in a ring $R$.

A Laurent polynomial ring is the localization of a ring $k[X_1,\dots ,X_r]$ obtained by inverting all variables. Its elements are finite sums of Laurent polynomials with exponent vectors in ${\mathbb{Z}}^r$. It is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring.

A nonzero quotient of an $r$-variable Laurent polynomial ring is finite over an embedded Laurent polynomial ring in $m\le r$ variables. Given a nonzero Laurent relation, choose $N$ so that weights $(N,N^2,\dots ,N^{r-1},1)$ separate its finite exponent support. The displayed invertible monomial substitution makes its extreme $Y_r$ coefficients units in the remaining Laurent variables. Normalize the relation to a monic polynomial with unit constant term; both $Y_r$ and $Y_r^{-1}$ are integral over the image of the $(r-1)$-variable subring. Induction and transitivity of finite extensions prove the assertion, over finite fields as well as infinite fields. An infinite-dimensional quotient has $m\ge 1$.

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 $A$.

Given a finite set $F\subset {\mathbb{Z}}^r$ of exponent vectors, there is $w\in {\mathbb{Z}}^r$ for which the integers $w\cdot m$, $m\in F$, are pairwise distinct. For $r=2$, take $w=(1,N)$ with $N$ larger than every first-coordinate difference in $F$. In higher dimensions take $w=(1,N,\dots ,N^{r-1})$ with $N$ sufficiently large. This proves that a nonzero Laurent polynomial can be detected by a substitution $X_i=T^{w_i}$ without cancellation of its distinct terms.

For an integral domain $R$, an integer-valued polynomial is a polynomial over its fraction field that maps $R$ into itself. Its coefficients need not lie in $R$: $X(X-1)/2$ maps $\mathbb{Z}$ into $\mathbb{Z}$. These polynomials form a ring containing $R[X]$.

For $n\ge 0$, define $B_0(X)=1$ and $B_n(X)=X(X-1)\cdots (X-n+1)/n!$. The binomial coefficient formula shows $B_n(m)\in \mathbb{Z}$ for every nonnegative integer $m$. Since these integers are dense in the p-adic integers, continuity of the polynomial implies $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$, so it is an integer-valued polynomial there and $\Vert B_n{\Vert }_{\infty }=1$. The Pascal's identity gives $\Delta B_{n+1}=B_n$ for the forward difference operator.

An automorphism of a polynomial ring is an invertible ring homomorphism from the ring to itself. For example, $f(X)\mapsto f(X+a)$ has inverse $f(X)\mapsto f(X-a)$, so translating the variable preserves reducibility.

Coefficientwise reduction defines a ring homomorphismIt preserves sums, products, and divisibility. The Frobenius endomorphism gives $\overline{g(X^p)}=\bar{g}(X{)}^p$.

If $R$ is an integral domain, the units of $R[x]$ are precisely the constant polynomials whose values are units of $R$.

A polynomial $f\in R[t]$ is a zero divisor exactly when some nonzero scalar $a\in R$ annihilates it. One proof chooses a nonzero polynomial $g$ of minimal degree with $fg=0$ and successively kills its leading coefficient.