transcendence degree over polynomial ring

This is Exercise 11 in Section 16.1 in Dummit&Foote's Abstract Algebra.

Let $V$ be an affine variety over a field $k$ and let $R = k[V]$ be its coordinate ring. Let $d_t(R)$ denote the transcendence degree of the field of fractions $k(V)$ over $k$, and let $d_p(R)$ be the Krull dimension of $R$ defined in terms of chains of prime ideals. This exercise shows $d_t(R) = d_p(R)$. By Noether's Normalization Lemma there is a polynomial subring $R_1 = k[y_1, \ldots, y_m]$ of $R$ such that $R$ is integral over $R_1$.

(a) Show that $d_t(R_1) = d_t(R) = m$ and that $d_p(R_1) = d_p(R)$. Deduce that we may assume $R = R_1$. [Use the Going-up and Going-down Theorems (cf. Theorem 26, Section 15.3) to prove the second equality.]

(b) When $R = R_1$ show that $d_p(R) \geq d_t(R)$ by exhibiting an explicit chain of prime ideals of length $m$.

(c) When $R = R_1$ show that any nonzero prime ideal of $R$ contains an element $f$ such that $R(f)$ is transcendental over $R$ of transcendence degree $1$. Use induction to show that $d_p(R) \leq d_t(R)$, and deduce that $d_p(R) = d_t(R)$.

For (c), the statement "$R(f)$ is transcendental over $R$ of transcendence degree $1$" confuses me a lot: what does $R(f)$ means? $R_f$ or $R/(f)$? And $R=R_1=k[y_1,\cdots,y_m]$ is not a field in general, so how can we define the transcendence degree over $R$?