Polynomials

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 5 November 2011

Polynomials

Let $𝔽$ be a field. If $a_0,a_1,a_2,\dots \in 𝔽$ use the notation $$a_0+a_{1}x+a_2{x}^2+\dots =\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i.$$

The polynomial ring $𝔽\left[x\right]$ is the set $$𝔽\left[x\right]=\left\{\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right|a_i\in 𝔽\text{and all but a finite number of the}{a}_i\text{are equal to}0\}$$ with operations given by $$\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right)+\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{b}_i{x}^i\right)=\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}(a_i+b_i)x^i\right)$$ and $$\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right)\left(\sum _{j\in {\mathbb{Z}}_{\ge 0}}{b}_j{x}^j\right)=\left(\sum _{k\in {\mathbb{Z}}_{\ge 0}}{c}_k{x}^k\right),\text{where}{c}_k=\sum _{i+j=k}{a}_i{b}_j.$$

The degree function is $\mathrm{deg}:𝔽\left[x\right]\to {\mathbb{Z}}_{\ge 0}$, where $$\mathrm{deg}(p_0+p_{1}x+p_2{x}^2+\dots )\text{is the maximal nonnegative integer}d\text{such that}{p}_d\ne 0.$$

Let $a\in 𝔽$. The evaluation homomorphism is $$\begin{array}{rcl}{\mathrm{ev}}_a:𝔽\left[x\right]& ⟶& 𝔽\\ p\left(x\right)& ⟼& p\left(a\right),\end{array}$$ where $$p\left(a\right)=p_0+p_{1}a+p_2{a}^2+\dots \text{if}p\left(x\right)=p_0+p_{1}x+p_2{x}^2+\dots .$$

The ring of formal power series in $x$ is $$𝔽\left[\right[x\left]\right]=\left\{\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right|a_i\in 𝔽\},$$ with operations given by $$\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right)+\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{b}_i{x}^i\right)=\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}(a_i+b_i)x^i\right)$$ and $$\left(\sum _{i\in {\mathbb{Z}}_{\ge 0}}{a}_i{x}^i\right)\left(\sum _{j\in {\mathbb{Z}}_{\ge 0}}{b}_j{x}^j\right)=\left(\sum _{k\in {\mathbb{Z}}_{\ge 0}}{c}_k{x}^k\right),\text{where}{c}_k=\sum _{i+j=k}{a}_i{b}_j.$$

Examples. The following are elements of $𝔽\left[\right[x\left]\right]$:

	${\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+\cdots ,}$
	${\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =\sum _{i\in {\mathbb{Z}}_{\ge 0}}\frac{x^i}{i!},}$
	${\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum _{i\in {\mathbb{Z}}_{\ge 0}}{(-1)}^i\frac{x^{(2i+1)}}{(2i+1)!},}$
	${\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots =\sum _{i\in {\mathbb{Z}}_{\ge 0}}{(-1)}^i\frac{x^{2i}}{\left(2i\right)!},}$
	${\displaystyle \ln (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots =\sum _{i\in {\mathbb{Z}}_{>0}}\frac{x^i}{i}.}$

1. $𝔽\left[x\right]$ is an integral domain.
2. $𝔽\left[\right[x\left]\right]$ is an integral domain.

HW: Show that $𝔽\left[x\right]$ with the degree function $\mathrm{deg}$ is a Euclidean domain.

The field of fractions of $𝔽\left[x\right]$ is the set $$𝔽\left(x\right)=\left\{\frac{a\left(x\right)}{b\left(x\right)}\right|a\left(x\right),b\left(x\right)\in 𝔽\left[x\right],b\left(x\right)\ne 0\}$$ with $$\frac{a\left(x\right)}{b\left(x\right)}=\frac{c\left(x\right)}{d\left(x\right)},\text{if}a\left(x\right)d\left(x\right)=b\left(x\right)c\left(x\right),$$ and with operations given by $$\frac{a\left(x\right)}{b\left(x\right)}+\frac{c\left(x\right)}{d\left(x\right)}=\frac{a\left(x\right)d\left(x\right)+b\left(x\right)c\left(x\right)}{b\left(x\right)d\left(x\right)}\text{and}\frac{a\left(x\right)}{b\left(x\right)}\cdot \frac{c\left(x\right)}{d\left(x\right)}=\frac{a\left(x\right)c\left(x\right)}{b\left(x\right)d\left(x\right)}.$$

The field of fractions of $𝔽\left[\right[x\left]\right]$ is the set $$𝔽\left(\right(x\left)\right)=\left\{\frac{a\left(x\right)}{b\left(x\right)}\right|a\left(x\right),b\left(x\right)\in 𝔽\left[\right[x\left]\right],b\left(x\right)\ne 0\}$$ with $$\frac{a\left(x\right)}{b\left(x\right)}=\frac{c\left(x\right)}{d\left(x\right)},\text{if}a\left(x\right)d\left(x\right)=b\left(x\right)c\left(x\right),$$ and with operations given by $$\frac{a\left(x\right)}{b\left(x\right)}+\frac{c\left(x\right)}{d\left(x\right)}=\frac{a\left(x\right)d\left(x\right)+b\left(x\right)c\left(x\right)}{b\left(x\right)d\left(x\right)}\text{and}\frac{a\left(x\right)}{b\left(x\right)}\cdot \frac{c\left(x\right)}{d\left(x\right)}=\frac{a\left(x\right)c\left(x\right)}{b\left(x\right)d\left(x\right)}.$$

1. The invertible elements of $𝔽\left[x\right]$ are the invertible elements of $𝔽$.
2. The invertible elements of $𝔽\left[\right[x\left]\right]$ are $a_0+a_{1}x+a_2{x}^2+\cdots \in 𝔽\left[\right[x\left]\right]$ with $a_0$ invertible in $𝔽$.

$𝔽\left(\right(x\left)\right)=\left\{x^{k}p\right(x\left)\right|k\in \mathbb{Z},p\left(x\right)\in 𝔽\left[\right[x\left]\right],p_0\ne 0\}\cup \left\{0\right\}$.

Let $p\left(x\right)\in 𝔽\left(\right(x\left)\right)$. The order $\nu \left(p\right)$ of $p\left(x\right)=\sum _{l\in \mathbb{Z}}{p}_l{x}^l$ is the minimal integer $l$ such that $p_l\ne 0$.

HW: Show that the order function $\nu :𝔽\left(\right(x\left)\right)\to \mathbb{Z}$ is a normalized discrete valuation (see [BouC] Ch. VI §3 no.6 def.3).

Notes and References

The polynomial ring is the universal commutative ring generated by a set with one element. It shares many properties with the integers $\mathbb{Z}$, the universal commutative group generated by a set with one element. PUT IN SOME REFERENCE TO BOURBAKI.

References

[BouC] N. Bourbaki, Commutative algebra, Masson, Hermann ??? MR???????? (20??e:20????)

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