For instance we know that $\mathrm{Frac}(\mathbb{Z}) = \mathbb{Q}$ or that $\mathrm{Frac}(2\mathbb{Z}) = \mathbb{Q}$ and we know that $\mathbb{Z} \subset \mathbb{Q}$.
These are fields, but do they also pass on their properties? $\mathbb{Z}$ doesn’t have an inverse for multiplication, but everything else is the same as $\mathbb{Q}$.
This answer unfortunately gets fairly technical fairly quickly; I've tried to keep it readable nonetheless, but please let me know if you have ideas for improvements. Also, per the comments I'm assuming that "field" is a typo for "ring."
Every equational sentence is preserved by the field-of-fractions construction. An equational sentence is an expression of the form $$\forall a_1,...,a_n (t(a_1,...,a_n)=s(a_1,...,a_n))$$ for terms $t,s$ in the relevant language. In the context of rings, "terms" are just polynomials - so for example the property of being a Boolean ring is expressible by an equational sentence, namely $\forall a(a*a=a)$. Equational expressions are quite limited, but there's still a lot of complexity there. The big theorem I'll use is the following:
$(*)\quad$ Suppose $\mathbb{A}$ is a quotient of a substructure of a (possibly infinitary) direct power of $\mathbb{B}$. Then every equational sentence true in $\mathbb{B}$ is true in $\mathbb{A}$ as well. And if $\mathbb{B}$ is (isomorphic to) a substructure of $\mathbb{A}$ as well, then the converse holds and $\mathbb{A}$ and $\mathbb{B}$ satisfy exactly the same equational sentences.
This is a special case of the easy direction of Birkhoff's HSP theorem (see the wiki page, or this old answer of mine if you're interested in an outline of the harder direction in the special case of groups), which basically showed that the operations of taking products, substructures, and homomorphic images (also called quotients) are the precise "basic ingredients" needed for equational-sentence-preserving constructions. I'll leave the proof of $(*)$ as an exercise for now; it's not especially hard, but it is somewhat tedious.
Let's suppose $R$ is an integral domain, that is, a commutative ring (with unity) and no zero divisors (see here for why that is a reasonable restriction). Its field of fractions $F_R$ can be defined as follows:
First we form the multivariable polynomial ring $R[\{x_r: r\in R\setminus\{0\}\}]$. I'll abbreviate that set of variables by "$X$" for simplicity.
Next, we mod out by the relations $x_rr=1$ for each $r\in R\setminus\{0\}$.
OK, so this gives two separate constructions: $R\leadsto R[X]$ ("adjoin lots of indeterminates") and $R[X]\leadsto F_R$ ("mod out by a messy ideal"). The latter is simply a surjective homomorphism and so preserves equations by $(*)$. But how do we adjoin indeterminates? Even figuring out how to adjoin a single indeterminate using just power, substructure, and quotient constructions is not obvious, but it can be done. To keep things concrete, let's think about the specific case of $R=\mathbb{Z}$. Consider the really big ring $\mathbb{Z}^\infty$ (I'm being a bit sloppy with notation here in the interest of readability) consisting of all infinite sequences of integers $(z_i)_{i\in\mathbb{N}}$ with componentwise addition and multiplication. There's an obvious way to see $\mathbb{Z}$ "sitting inside" $\mathbb{Z}^\infty$, namely by considering the map $$\eta:\mathbb{Z}\rightarrow \mathbb{Z}^\infty: z\mapsto (z,z,z,z,...).$$ On the other hand, we also have wacky things like the identity sequence $$\iota=(1,2,3,4,...)$$ (I'm starting indexing from $1$ instead of $0$ here). The point is that such "wacky" elements of $\mathbb{Z}^\infty$ behave like indeterminates in the sense that (for example) if we write "$\langle U\rangle_W$" for the subring of $W$ generated by $U$ there is an isomorphism $$\mathbb{Z}[x]\cong \langle ran(\eta)\cup\{\iota\}\rangle_{\mathbb{Z}^\infty}.$$ This tells us that $\mathbb{Z}[x]$ - up to isomorphism, anyways - is indeed a subring of a power of $\mathbb{Z}$ and so satisfies all the equational sentences that $\mathbb{Z}$ itself does. Of course we're not done - we've only figured out how to adjoin a single indeterminate to the specific ring $\mathbb{Z}$ - but the broader result we need is no harder to get.
Incidentally, this same argument gives as a special case that the fields $\mathbb{R}$ and $\mathbb{C}$ have the same equational theory; see the latter part of this old answer of mine for more on this. Incidentally, this specific fact can also be proved using complex analysis (think about what happens if a polynomial vanishes on all real inputs).
