8.2: Campo de Fracciones

En la historia de los sistemas numéricos, hay una clara progresión: Ante un vacío donde podría haber más números, se inventan más números. Primero vinieron los números naturales (los números de conteo), y cuando la gente encontró que ciertos problemas de resta no tenían solución, se introdujeron números negativos para llenar el vacío. Relevante para nuestra discusión actual, los números racionales surgen cuando uno nota que algunos problemas de división entre enteros no tienen solución.

Supongamos que\(R\) is an integral domain. Then the only impediment to division is a lack of actual quotients: if the quotients were to exist, they would have to be unique.

Considerar\(\mathbb{Z}\). This is of course an integral domain, but wouldn't it be nice if \(2\) had a multiplicative inverse? We'll extend \(\mathbb{Z}\) by including \(\frac{1}{2}\). But when we include \(\frac{1}{2}\), we also have to include all possible sums and products in order to ensure that we still have a ring; the operations of addition and multiplication need to be closed, of course. So in addition to \(\frac{1}{2}\), we also need to include every number \(\frac{n}{2^m}\), with \(n\in \mathbb{Z}\) and \(m\in \mathbb{N}\) in order to ensure that the set is closed under multiplication and addition. Call this set \(R\). Then \(R\) is a commutative ring with unity. It's also an integral domain, but still not a field, since, for example, \(3\) has no multiplicative inverse.

En ese caso, podemos seguir adelante e incluir el inverso multiplicativo de cada entero positivo, junto con todas las sumas y productos posibles de esos inversos. El anillo resultante, por supuesto, son los números racionales,\(\mathbb{Q}\).

Nos gustaría extender esta construcción a un dominio integral arbitrario: Partiendo de un dominio integral\(D\), we introduce inverses and the appropriate sums and products until every element has an inverse. In fact, this involves copying the whole notion of fractions.

Primero construimos un anillo\(D'\). For an integral domain \(D\), \(D'\) is the set \(D\times (D\setminus \{0\})\). Each pair \((a,b)\) in \(D'\) can be thought of as fractions \(\frac{a}{b}\); note that we disallow \(b=0\). The operation \(+\) defined by \((a,b)+(x,y) = (ay+bx, by)\) and multiplication is defined by \((a,b)\cdot (x,y) = (ax, by)\). Then \(D'\) is a commutative ring; we leave it as an exercise to show this is true.

Let\(D\) be an integral domain. Show that \(D'\) is an integral domain. (In particular, check all of the ring axioms, and then show that there are no zero-divisors in \(D'\).)

Con números racionales, es importante notar que muchas fracciones diferentes son iguales: actualmente nuestro anillo\(D'\) is much too large! For example, we haven't introduced any mechanism for cancellation of numerator and denominator: \((a,b)\cdot (b,a)=(ab,ab)\neq 1\) in \(D'\).

Construiremos el anillo real de fracciones como un cociente de\(D'\). To construct a quotient, we only need to identify a suitable ideal \(I\); the quotient will then just be \(D'/\mathord I\). The ideal should contain everything that is 'equivalent to \(0\)' in the ring of fractions. Thinking by analogy to the rationals, we see that this is the set \(I=\{(0,x) \mid x\in D\}\).

Este conjunto se muestra fácilmente como un ideal:\((0,a)+(0,b)=(0ab, ab)=(0,ab)\in I\). And for any \((x,y)\in D'\), we have \((x,y)\cdot (0,a)=(0,ya)\in I\). Then \(I\) is an ideal.

Ahora podemos comprobar que\((a,b)\cdot (b,a)=(1,1)\) in the quotient \(D'/\mathord I\): \((a,b)\cdot (b,a)=(ab, ab)\), and \((ab,ab)-(1,1)=(ab-ab,ab)=(0,ab)=0\), as desired.

Luego definimos el campo de fracciones como\(Q=D'/\mathord I\). This is in fact a field: For any \(a,b \neq 0\), we have \((a,b)^{-1}=(b,a)\), so every non-zero element in \(Q\) has a multiplicative inverse.

Definición 8.1.1: Campo de Fracciones

El campo de fracciones de un dominio integral\(D\) is \(D'/\mathord I\), with \(D'\) and \(I\) as defined above.

Contributors and Attributions

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