Theorem. Let $Z\sim N(0,1)$ and $Y\sim \chi^2(v)$ are two independent random variabels. Then, $$T = \frac{Z}{\sqrt{Y/v}} \sim t-student$$ with degrees of freedom $v$.
Is it the converse always true? If $T\sim t-student$ with degrees of freedom $v$, then there exist two random variables $Z\sim N(0,1),Y\sim \chi^2(v)$ independently such that $T=\frac{Z}{\sqrt{Y/v}}$?
I guess we do not need to worry about the existence of such a random variable. They must exist, and you know what they are. Instead, we should treat $\dfrac{Z}{\sqrt{Y/v}}$ and $T_v$ distribution as the same thing; it is just a different representation of the same distribution.
There are two other examples of random variables with different representations.
Discrete Random Variable: We know the sum of $n$ iid Bernoulli$(p)$ random variables is following a binomial$(n, p)$ distribution. Therefore, we can always write $X\sim Bin(n,p)$ as $$X=\sum_{i=1}^n Y_i$$, where $Y_i \overset{i.i.d}{\sim} Bernonuli(p)$.
Continuous Random Variable: For $X\sim N(\mu,\sigma^2)$, we can always write it as $$X = \mu + \sigma Z$$, where $Z\sim N(0,1)$.