## Prime ideals and rings of fractions

The concept of prime ideal in a commutative ring $R$ with $1$ is one of several natural generalisations of the concept of prime integer number.

An ideal $I\subset R$ is called prime if for any $u,v\in R$ the following holds:
$uv\in I$ implies that at least one of these elements, $u$ and $v,$ is in $I.$

E.g. the principal ideal $(p)\subset \mathbb{Z}$ is prime if and only if $p$ is prime.

A maximal ideal $I\subset R$ is prime.

Indeed, let $uv\in I.$ Assume that $u\not\in I.$ We need to show that then $v\in I.$ If this were not the case then $u+I$ and $v+I$ are two non-zero elements in the ring $R/I$ such that $(u+I)(v+I)\subseteq I.$ But this is not possible, as $R/I$ is a field. Thus $v\in I,$ as claimed.

Analysing this proof, one can easily see that

If $I\subset R$ is prime then $R/I$ has no zero-divisors, i.e. it is an integral domain.

Further important property of prime ideals is that they are radical, i.e.
$I=\sqrt{I},$ where the radical $\sqrt{I}$ of the ideal $I$ ideal is $\sqrt{I}=\{x\in R\mid \exists k\geq 1 : x^k\in I\}.$ Indeed, $x^k\in I$ implies that either $x$ or $x^{k-1}$ is in $I,$ and we derive $x\in I$ by applying this reduction.

Yet another interesting observation is that $R\setminus I$ is multiplicatively closed.

Rings of fractions

A subset $S\subset R$ is called multiplicatively closed (or multiplicative) if $0\not\in S,$ $1\in S,$ and $uv\in S$ for any $u,v\in S.$

Given a multiplicatively closed set $S\subset R,$ one defines
a relation on $R\times S,$ as follows:

${\displaystyle (y,t)\equiv (x,s)\quad\text{iff there exists }u\in S: uys=uxt.}$

It is not hard to show that $\equiv$ is an equivalence relation.
To simplify the notation, write its equivalence class with representative $(x,s)$ as $\frac{x}{s}.$ We define

$S^{-1}R=(R\times S)/\equiv$ is the ring of fractions of $R$ w.r.t. $S,$ with addition and multiplication given by the rules
$\frac{x}{s}+\frac{y}{t}=\frac{xt+ys}{st},$ $\frac{x}{s}\cdot\frac{y}{t}=\frac{xy}{st}.$

It is easy to check that this is well-defined. We also have

$\phi: R\to S^{-1}R,$ so that $\phi(x)=\frac{x}{1},$ is a ring homomorphism.

Note that $\phi$ need not be injective, i.e. $\phi(R)\cong R$ need not hold. Indeed, if $x\in R$ is a zero-divisor such that $xu=0$ for $u\in S$ then $(x,1)\equiv (0,1),$ and so $\phi(x)=\frac{x}{1}=\frac{0}{1}=0_{S^{-1}R}.$

The most well-known example is the case $R$ being an integral domain, and $S=R\setminus\{0\}.$ Then $S^{-1}R$ is a field, called the field of fractions of $R.$

Examples:

• $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$
• for the ring of polynomials $\mathbb{F}[T]$ over a field $\mathbb{F}$ the field $\mathbb{F}(T)$ is the field of rational functions over $\mathbb{F} .$
• Let $x\in R$ be non-nilpotent. Then $S=\{1,x,x^2,x^3,\dots\}$ isa multiplicative set. Moreover, then $S^{-1}R\cong R[T]/(Tx-1)$ (it is not completely trivial to prove this, though). Intuitively, we make the variable $T$ behave like the inverse of $x,$ as $Tx=1$ in this ring.

Now let us look at the case $S=R\setminus I,$ for $I$ a nonzero prime ideal. In this case $S^{-1}R$ is denoted by $R_I$ and called the localisation of $R$ at $I.$

Example: Let $\mathbb{F}[T]$ be the ring of polynomials over a field $\mathbb{F},$ and $a\in \mathbb{F}.$ Then $I=(T-a)$ is prime, and $R_I=\mathbb{F}[T]_{(T-a)}$ is equal to $\{\frac{f}{g}\in \mathbb{F}(T)\mid (T-a)\not|\, g \}.$

The ring $R_I$ has unique maximal ideal $IR_I.$

It suffices to show that $\frac{x}{s}$ is invertible in $R_I$ iff ${x}\not\in I.$ Indeed, if $x\in S$ then $\frac{x}{s}\frac{s}{x}=1.$ On the other hand, if $\frac{x}{s}\frac{y}{t}=1$ then there exists $u\in S$ such that $uxy=ust.$ Thus $uxy\not\in I,$ and so $x\not\in I$ (if it was, $uxy$ would be in $I,$ as $I$ is an ideal).