# Finding the field of fractions of a quotient of a polynomial ring?

This should be very basic but I am having a bit of trouble finding the field of fractions for quotients of polynomial rings over a field. The specific example I am having trouble with is the following: Let $$k$$ be a field with characteristic $$\neq 2$$ and let $$f \in k[x_1, x_2, \ldots , x_n]$$ be a polynomial with no repeated factors. I am trying to determine the field of fractions of the ring $$k[x_1, x_2, \ldots , x_n, z]/\langle z^2 - f\rangle.$$ The usual method I would use for a problem like this is the Chinese remainder theorem, but I don't see any way to apply that here. The other fact I know is that if $$A$$ is an integral domain with prime ideal $$\mathfrak{p}$$, then $$\text{Frac}(A/\mathfrak{p}) \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$. But the problem of finding $$A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$ seems no easier in this case. Can anyone point me in the right direction?

I am fairly certain I know what the result is, but I have no idea how to show it properly.

• Isn't the field of fractions just $k(x_1, \ldots, x_n)[z]/(z^2 - f)$? It's certainly a field containing your ring, and i think you can argue that it is the smallest such. – André 3000 Feb 13 at 5:39
• @André3000 that is the field of fractions, yes, but could you explain how exactly you'd show that? – Joe Feb 15 at 18:17

## 1 Answer

Let $$R = k[x_1, x_2, \ldots , x_n, z]/(z^2 - f) = k[x_1, \ldots, x_n][\overline{z}]$$, let $$F = \operatorname{Frac}(R)$$ be its field of fractions, and let $$K = k(x_1, \ldots, x_n)[z]/(z^2 - f) = k(x_1, \ldots, x_n)[\overline{z}]$$. Since $$K$$ is a field containing $$R$$ and $$F$$ is the smallest such field, then we have $$R \subseteq F \subseteq K$$. Now $$F$$ is a field containing $$x_1, \ldots, x_n$$, so it must also contain $$k(x_1, \ldots, x_n)$$ and since $$F$$ also contains $$\overline{z}$$ (the image of $$z$$ in the quotient), then $$K \subseteq F$$.