# Definition of a valuation of a rational fraction

Let $$F$$ be a field. Then the field of rational fractions over $$F$$ in indeterminates $$x_1,...,x_n$$, denoted by $$F(x_1,...,x_n)$$, is the field of fractions of the polynomial ring $$F[x_1,...,x_n]$$.

P.A.Grillet ("Abstract Algebra") defines a valuation of a rational fraction $$f(x_1,...,x_n)/g(x_1,...,x_n)$$ at $$(a_1,...,a_n) \in F^n$$ as $$f(a_1,...,a_n)g(a_1,...,a_n)^{-1}$$, but he says it is only defined if $$g(a_1,...,a_n) \neq 0$$ and in this case it invariant under the change of the representative of the equivalence class which is the rational fraction. Concretely, if $$f_1(x_1,...,x_n), f_2(x_1,...,x_n), g_1(x_1,...,x_n), g_2(x_1,...,x_n)$$ are polynomials over $$F$$ so that $$g_1$$ and $$g_2$$ are nonzero and $$f_1(x_1,...,x_n)g_2(x_1,...,x_n) = f_2(x_1,...,x_n)g_1(x_1,...,x_n)$$, then $$f_1(a_1,...,a_n)g_1(a_1,...,a_n)^{-1} = f_2(a_1,...,a_n)g_2(a_1,...,a_n)^{-1}$$.

I understand why $$f_1(a_1,...,a_n)g_1(a_1,...,a_n)^{-1} = f_2(a_1,...,a_n)g_2(a_1,...,a_n)^{-1}$$ if $$g_1(a_1,...,a_n), g_2(a_1,...,a_n) \neq 0$$ and $$f_1(x_1,...,x_n)g_2(x_1,...,x_n) = f_2(x_1,...,x_n)g_1(x_1,...,x_n)$$ (it follows directly from the fact that evaluation $$F[x_1,...,x_n]\to F$$ at $$(r_1,...,r_n)$$ is a homomorphism of rings.

What I don't understand if why we necessarily have $$g_2(a_1,...,a_n) \neq 0$$ if $$g_1(a_1,...,a_n) \neq 0$$ and $$f_1(x_1,...,x_n)g_2(x_1,...,x_n) = f_2(x_1,...,x_n)g_1(x_1,...,x_n)$$. Indeed, if $$g_1(a_1,...,a_n) \neq 0$$ it is possible that $$g_2(a_1,...,a_n) = 0$$ together with $$f_2(a_1,...,a_n) = 0$$.

What I don't understand if why we necessarily have $$g_2(a_1,...,a_n) \neq 0$$ if $$g_1(a_1,...,a_n) \neq 0$$ and $$f_1(x_1,...,x_n)g_2(x_1,...,x_n) = f_2(x_1,...,x_n)g_1(x_1,...,x_n)$$.

No, that's not true, so there's no way to prove it. To see this, take a particular example.

Let $$g_1(x) = x+8964$$, $$g_2(x) = (x+8964)(x + 1997)$$, $$f_1(x) = x+1$$, $$f_2(x) = (x+1)(x + 1997)$$. It's easy to see that $$f_1(x)g_2(x) = f_2(x)g_1(x)$$ and $$g_1(-1997) \ne 0$$, but $$g_2(-1997) = 0$$.

• Hmm, I suspected this may be a case. Does this mean that the evaluation of rational fractions is not well-defined and useless? – Jxt921 Mar 4 at 10:04
• @Jxt921 No, the definition is good. You simply need $g_2(a)^{-1} \ne 0$ when you write $f_1(a_1,...,a_n)g_1(a_1,...,a_n)^{-1} = f_2(a_1,...,a_n)g_2(a_1,...,a_n)^{-1}$. – GNUSupporter 8964民主女神 地下教會 Mar 4 at 10:07
• @Jxt921 I mean you only need to include $g_i(a_1,\dots,a_n) \ne 0$, $i = 1,2$ into the assumptions of the definition. – GNUSupporter 8964民主女神 地下教會 Mar 4 at 10:17

Well, consider the general construction of the field of fractions from an integral domain $$R$$.

For this, one defines an equivalence relation on $$R\times (R\setminus\{0\})$$: $$(a,b) \equiv (c,d)$$ if $$ad=bc$$. The equivalence classes are written as fractions: $$a/b := \{(c,d)\mid (a,b)\equiv (c,d)\}$$. On the quotient set $$K =\{a/b\mid a,b\in R,b\ne 0\}$$ define addition and multiplication as usual. In this way, $$K$$ becomes a field and $$R\rightarrow K:r\mapsto r/1$$ is an embedding (ring monomorphism).

The construction shows that for the fractions $$a/b$$ one has $$b\ne 0$$ by definition.

• OP is not sure about the possibility that $b \in R \setminus \{0\}$ is mapped to $0_R$ by an evaluation homomorphism $g_2(a_1,\dots,a_n)$ for some polynomial $g_2$ and $a_1,\dots,a_n \in K$. How does your answer address that? – GNUSupporter 8964民主女神 地下教會 Mar 4 at 10:12