Today we discuss something on polynomials.

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Over a Commutative Ring

Nilpotents and units are closely related. In a commutative unital ring $R$, if $x$ nilpotent, $a$ unit, then $a+x$ is again a unit. If $1+x y$ is a unit for every $y\in R$, then $x\in\mathfrak{R}$, the Jacobson radical, approximately nilpotent.

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 $f g=a_0 b_0 + (a_0 b_1+a_1 b_0)x+(a_0 b_2+a_1 b_1+a_2 b_0)x^2+...$ 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.