Today we discuss something on polynomials.

Over a Commutative Ring

Let $A$ be a commutative unital ring, and $A[x]$ the polynomial ring over $A$.
Let $f=a_0+a_1x+...+a_n x^n$. If $a_1,a_2,...,a_n$ are nilpotent, so will be $f-a_0$. If moreover $a_0$ is invertible, $f$ will be invertible; if instead $a_0$ is nilpotent, $f$ is nilpotent. The converses are both true. For nilpotency, the highest degree term of $f^m$ is a sole $a_n^m x^m$, if $f$ is nilpotent, $a_n$ is forced to be; but then $f-a_n x^n$ is again nilpotent. For invertibility, immediately $a_0$ is invertible; Suppose $fg=1$ with $g=b_0+b_1x+...+b_r x^r$. Then $a_n b_r=0,a_n b_{r-1}+a_{n-1} b_r=0,...$. Multiplying the second by $a_n$, we get $a_n^2 b_{r-1}=0$; repeating this yields $a_n^{r+1} b_0=0$, and $b_0$ is invertible so $a_n$ is nilpotent.

In particular, these implies the nilradical $\mathfrak{N}=\mathfrak{R}$ in polynomial rings. If $f\in\mathfrak{R}$, then $1+xf$ is invertible. This means $a_0,...,a_n$ are all nilpotent, hence $f$ nilpotent. In the proof of the Hilbert Nullstellensatz, we will see that this is valid also in prime quotients of polynomial rings.

If $f$ is a zero-divisor, then $a_0,..,a_n$ are all zero-divisors. Indeed, if $fg=0$, then $a_n b_r=0$, and $f a_n g =0$, with $\mathrm{deg} a_n g<\mathrm{deg} g$. Repeating this, eventually $a_n g$=0. This yields $(f-a_n x^n) g=0$. Then $a_i g=0,a_i b_n=0,\forall i$.

A general version of Gauss’s lemma holds: if $(a_0,...,a_n)=(1)$, then $f$ is said to be primitive. If $f,g$ are primitive, then so is $f g$. The proof is analogous: If $(c_0,...,c_n)\in\mathfrak{p}$ for some maximal $p$, then in $(A/\mathfrak{p}[x]$, we have $f g=0$. Since this is a domain, either $f,g$ is $0$, a contradiction.

The above is easily generalized to several variables (actually arbitrarily many, since a polynomial always involves only finite terms), keeping in mind $A[X_1,...,X_n]=A[X_1,...,X_{n-1}][X_n]$.

The case of power series is different in many aspects. First, if $f=a_0+a_1 x+...$, then $f$ is invertible if and only if $a_0$ is. This is because suppose $g=b_0+b_1 x+...$, then KaTeX parse error: Expected 'EOF', got ' ' at position 12: f g=a_0 b_0 ̲+ (a_0 b_1+a_1 … where $a_i$ can be solved inductively as long as $a_0 b_0=1$. Second, although $f$ nilpotent implies $a_i$ nilpotent for all $i$, via some similar induction focusing on the lowest degree term, the converse is not true. In fact, there are some restrictions on the vanishing degree: if $f^s=0$, then $a_0^s=0$, so $(f-a_0)^{2s}=0$; then $a_1^{2s}=0$, so $(f-a_1 x)^{4s}=0$. In general $a_i^{2^i s}=0$. If the least $s_i$ for $a_i^{s_i}=0$ increases rapidly, making $2^{-i} s_i\rightarrow\infty,i\rightarrow \infty$, then $f$ is not nilpotent. For example take $s_i=3^i,A=\prod_{i\in\mathbb{Z}^+}\mathbb{C}[x_i]/(x_i^{s_i}),a_i=x_i$. The argument also applies in the polynomial case, but then $n$ is finite.

If $1+g f$ is invertible iff $1+a_0 b_0$ is invertible. So $f\in\mathfrak{R}(A[[x]])$ iff $a_0\in\mathfrak{R}(A)$.

The ideal $F(\mathfrak{I})$ of $f$ with $a_0\in \mathfrak{I}$ is an ideal of $A[[x]]$. Moreover $A/\mathfrak{I}\cong A[[x]]/F(\mathfrak{I})$. So if $\mathfrak{I}$ is prime, so is $F(\mathfrak{I})$; same for maximality. In fact, the same holds in $A[x]$.

The above topic is from Atiyah, M. F.; MacDonald, I. G. (February 21, 1994). «Chapter 1: Rings and Ideals». Introduction to Commutative Algebra. Westview Press. p. 11. ISBN 978-0-201-40751-8.

The case of countable variables is also of interest. We will discuss this in later posts.

Thursday, August 1, 2013