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.


  • \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).


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