## Polynomials

Let $𝔽$ be a field. If ${a}_{0},{a}_{1},{a}_{2},\dots \in 𝔽$ use the notation $a0+ a1x+ a2x2+ … = ∑i∈ ℤ≥0 aixi.$

The polynomial ring $𝔽\left[x\right]$ is the set $𝔽[x]= { ∑i∈ ℤ≥0 aixi | ai∈𝔽 and all but a finite number of the ai are equal to 0 }$ with operations given by $( ∑i∈ ℤ≥0 aixi ) + ( ∑i∈ ℤ≥0 bixi ) = ( ∑i∈ ℤ≥0 (ai+bi) xi )$ and $( ∑i∈ ℤ≥0 aixi ) ( ∑j∈ ℤ≥0 bjxj ) = ( ∑k∈ ℤ≥0 ckxk ) ,where ck= ∑i+j=k aibj .$

The degree function is $\mathrm{deg}:𝔽\left[x\right]\to {ℤ}_{\ge 0}$, where $deg(p0+ p1x+ p2x2+ …) is the maximal nonnegative integer d such that pd≠0 .$

Let $a\in 𝔽$. The evaluation homomorphism is $eva: 𝔽[x] ⟶ 𝔽 p(x) ⟼ p(a),$ where $p(a) = p0+ p1a+ p2a2+ … if p(x) = p0+ p1x+ p2x2+ … .$

The ring of formal power series in $x$ is $𝔽[[x]] = { ∑i∈ ℤ≥0 aixi | ai∈𝔽 } ,$ with operations given by $( ∑i∈ ℤ≥0 aixi ) + ( ∑i∈ ℤ≥0 bixi ) = ( ∑i∈ ℤ≥0 (ai+bi) xi )$ and $( ∑i∈ ℤ≥0 aixi ) ( ∑j∈ ℤ≥0 bjxj ) = ( ∑k∈ ℤ≥0 ckxk ) ,where ck= ∑i+j=k aibj .$

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

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

1. $𝔽\left[x\right]$ is an integral domain.
2. $𝔽\left[\left[x\right]\right]$ 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 $𝔽(x) = { a(x) b(x) | a(x), b(x) ∈𝔽[x], b(x)≠0 }$ with $a(x) b(x) = c(x) d(x) , if a(x) d(x) = b(x) c(x) ,$ and with operations given by $a(x) b(x) + c(x) d(x) = a(x)d(x) + b(x)c(x) b(x)d(x) and a(x) b(x) ⋅ c(x) d(x) = a(x)c(x) b(x)d(x) .$

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

1. The invertible elements of $𝔽\left[x\right]$ are the invertible elements of $𝔽$.
2. The invertible elements of $𝔽\left[\left[x\right]\right]$ are ${a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots \in 𝔽\left[\left[x\right]\right]$ with ${a}_{0}$ invertible in $𝔽$.

$𝔽\left(\left(x\right)\right)=\left\{{x}^{k}p\left(x\right)\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}k\in ℤ,\phantom{\rule{0.5em}{0ex}}p\left(x\right)\in 𝔽\left[\left[x\right]\right],\phantom{\rule{0.5em}{0ex}}{p}_{0}\ne 0\right\}\cup \left\{0\right\}$.

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

HW: Show that the order function $\nu :𝔽\left(\left(x\right)\right)\to ℤ$ 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 $ℤ$, 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????)