Why $\deg(f)\ge1$ is required in the hypothesis of this statement? There is this statement that I don't quite understand its hypothesis.
Let $R$ be a UFD and $F$ its field of quotients. If $f\in R[X]$ is irreducible in $R[X]$ and $\deg(f)\ge1$, then $f$ is irreducible in $F[X]$.
My questions are:
1. Why is $\deg(f)\ge1$ required in the hypothesis? Suppose $f$ is a constant, then isn't $f$ still irreducible?
2. When they talk about its field of quotients, which specific quotients are they mentioning? Is there only one unique field of quotients for each UFD?
I will appreciate if anyone could provide some explanations with examples.
Thanks!
 A: The requirement about the degree is to make sure the polynomial is not a unit (i.e., an invertible element in the $F[X]$. Units are generally not considered to be irreducible e.g., $2$ is irreducible in $\mathbb Z$ but not in $\mathbb Q$ since $2=4\cdot \frac{1}{2}$ ($2$ is a unit $\mathbb Q$). 
As for the second question, up to isomorphism there is exactly one field of quotients for a given $UFD$. You can find plenty of sources that will give constructions of the field of quotients including proofs of the essential uniqueness. 
A: Let's just check the definition of irreducible for an integral domain $R$:

A non-zero non-unit $x\in R$ is said to be irreducible if it is not the product of two non-units.

So $0$ and units can never be irreducible. For example: let $R = \Bbb R$. Then consider the constant polynomial $3\in \Bbb R[x]$. This is not irreducible, because $1/3\in \Bbb R[x]$, and $1/3\cdot 3 = 1$, so $3$ is a unit (in this way, all non-zero constant polynomials are units in $\Bbb R[x]$). Now, if $R$ is not a field, you might have a constant polynomial that is irreducible: for example, $2\in\Bbb Z[x]$. However, as we will see, this becomes a unit when we move to the field of quotients you mentioned.
For an integral domain $R$, we can construct its field of quotients (or field of fractions), which is defined abstractly as the smallest field $\operatorname{Frac}R$ in which $R$ can be embedded. More concretely, $$\operatorname{Frac}R = \left\{\frac{a}{b}\mid a,b\in R, b\neq 0\right\}.$$ If our original ring is a polynomial ring, say something like $\Bbb Z[x]$, an element of $\operatorname{Frac}\Bbb Z[x]$ will look like
$$
\frac{a_n x^n + \dots + a_1 x + a_0}{b_m x^m + \dots + b_1 x + b_0},
$$
where $a_i, b_i\in \Bbb Z$ and $b_m x^m + \dots + b_1 x + b_0$ is not the $0$ polynomial. So, elements of $\operatorname{Frac}\Bbb Z[x]$ are rational functions with coefficients in $\Bbb Z$. Another concrete example, even simpler: $\operatorname{Frac}\Bbb Z = \Bbb Q$. Here we see why we must exclude constant polynomials - although $2$ is irreducible in $\Bbb Z[x]$, $1/2\in\operatorname{Frac}\Bbb Q[x]$, so $2\cdot\frac{1}{2} = 1$, and $2$ is a unit in $\left(\operatorname{Frac}\Bbb Z\right)[x] = \Bbb Q[x]$.
In general, since you're looking just at some ring $R$, the field of fractions you'll get will be fractions of elements in $R$. As Ittay Weiss mentioned, there is one field of fractions for any UFD (up to isomorphism), so the answer to that part of your question is "yes."
