Fields of fractions

In my textbook there's a theorem that goes like this:

Let $D$ be an integral domain. Then $D$ can be embedded in a field of fractions $F_D$, where any element in $F_D$ can be expressed as the quotient of two elements in $D$. Furthermore, the field of fractions $F_D$ is unique in the sense that if $E$ is any field containing $D$, then there exists a map $\phi : F_D \rightarrow E$ giving an isomorphism with a subfield of $E$ such that $\phi (a) = a$ for all elements $a\in D$.

I don't understand what it means to say that $D$ can be embedded in the field of fractions. I just don't get what the word "embedded" means, my book doesn't define it!

And what does it mean to say that the field of fractions $F_D$ is unique? What are they getting at by saying that?

Also, I don't understand the "significance" of this theorem. It seems very random and out of the blue to me. Can anyone enlighten me?

Sorry if this is a stupid question, but I really don't understand this at the moment, despite having tried and tried for hours. What makes it worse is that I love abstract algebra and really want to master these concepts but at the moment I am not!

• Take $D=\mathbb{Z}$. It can be embedded into its field of fractions $Quot(\mathbb{Z})=\mathbb{Q}$, i.e., there is an injective ring homomorphism $\mathbb{Z}\hookrightarrow \mathbb{Q}$. – Dietrich Burde Jun 24 '15 at 14:57

Embedding, on this context, means there is an injective homomorphism (that is preseves structure) $\phi : D \to F_D$. For example take $D = \mathbb Z$ and $F_D = \mathbb Q$. You may consider \begin{align}\phi : \mathbb Z &\to \mathbb Q\\ a &\mapsto \frac{a}{1}\end{align}
Notice that $\mathbb Q$ is the fields of fraction of $\mathbb Z$ through the relation $$(a,b)\sim(a',b') \iff a\cdot b' = a' \cdot b$$
An integral domain $D$ is not a field, so one might ask what elements we need to introduce/adjoin to $D$ in order to make it a field. The theorem claims that the field of fractions $F_D$ (see your text for the definition) is not only a field that contains $D$, but it is also the "unique smallest" such field, since any other field containing $D$ will have some subfield that is isomorphic to $F_D$.
Saying that $X$ can be embedded in $Y$ means that there is a (natural) structure-preserving injection $X\to Y$. In your case, the embedding from $D\to F_D$ is the map \begin{equation*} d\mapsto \frac{d}{1}. \end{equation*} You can use the definition of $F_D$ together with the fact that $D$ is a domain to show that this is an injection.
Now, if $E\supseteq D$ is a field containing $D$, then every element of $D$ is invertible in $E$, so that $E$ also contains a subfield $F$ isomorphic to $F_D$. $F$ is just the smallest subfield of $E$ containing both $D$ and the inverses of all nonzero elements of $D$ in $E$. There is clearly a natural map from $F_D\to F$ given by $$\frac{a}{b}\mapsto ab^{-1}.$$ Thus $F_D$ is unique up to isomorphism.