Dimension of a curve I'm trying to understand this example:

Let $f(T_1,T_2)\subset k[T_1,T_2]$ be a non-constant irreducible polynomial. Let
  $X=Z(f)\subset \mathbb A^2$. We will see that $\dim(X)=1$. We have
  $k[X]=k[T_1,T_2]/(f)$ and 
$$\dim(X)= \operatorname{tr.deg}_k k(X) \lneq \operatorname{tr.deg}_k k(T_1,T_2) = 2.$$
Since in $k(X)$ the generators $T_1, T_2$ follow to an algebraic
  relation $f$. On the other hand, $\dim(X)\ge 1$ since $X$ is not finite, thus $\dim (X)=1.$

I didn't understand why the $\lt$ part and why $\operatorname{tr.deg}_k(k(T_1,T_2))=2$. Could someone help me?
Thanks a lot.

$T_1$ and $T_2$ satisfy an algebraic relation in $k[X] = k[T_1,T_2]/(f)$, but not in $k(T_1,T_2)$.  The latter notation is specifically chosen to refer to the field in which $T_1$ and $T_2$ are formal variables with no relations, and the transcendence degree of such a field over $k$ equals $2$ by pretty much any definition of transcendence degree.
