How to replace identifications with equalities? This is a question which has been bothering me a lot since I've began studying abstract mathematics. I'm not sure if others are also bothered by this, or if this is something not to worry about.
A lot of times, there is a statement about extending an object to a larger one, such that the first is a sub-object of the other. And a lot of places, we replace the object by an isomorphic (in some sense) copy of itself, which is not really the question, because we wanted to create a larger object containing the first as a subset.
For example, field extensions: if $F$ is a field and $p(x)$ is an irreducible polynomial over this field, then there is a field extension $K/F$ such that $p(x)$ has a root in $K$. A usual proof of this is to consider $F[x]/(p(x))$, where we identify $F$ with a copy of it in $F[x]/(p(x))$, which more or less gets the job done, but I'm still not able to convince myself that there is a field $K$ with the given property such that $F\subset K$.
There are many more examples of situations like this (I am not able to recall a lot of them), where people just work with isomorphic copies.
So my question is: are mathematicians fine to just work with isomorphic copies, and how in particular do I convince myself in the above question?
 A: This answer addresses the question of how to prove an extension exists which contains $F$ set-theoretically as a subfield.
Let $F'$ be the isomorphic image of $F$ sitting inside $K' = F[x]/(p(x))$ consisting of constants. Naïvely, one would like to simply replace the elements of $F'$ with elements of $F$. The problem is that it is not at all clear that there can't be elements of $K' - F'$ that also belong to $F$.
There are a few possible solutions I see to this objection.
First solution (using the axiom of foundation).
Use the set-theoretical axiom of foundation to show that this doesn't occur. Namely, say $f \in F$ also occurs in $K'$.
An element of $K'$ is an equivalence class $A$ in $F[x]$. But since $p(x)$ is nonzero, a representative $q(x) \in A$ can have any leading coefficient whatsoever, including $f$. (If $f = 0$, we consider instead the zero coefficient above the leading coefficient of $q$.) Now the polynomial $q(x)$ is set-theoretically a sequence of elements in $F$ one of whose members is $f$. Working through the definitions--things may vary slightly depending on your conventions--we get a sequence of the form
$$f \in u_1 \in u_2 \in \dots \in u_r \in q \in A,$$
where $r$ is a small integer that can be determined explicitly.
This contradicts the axiom of foundation since we're assuming $A = f$.
This solution is perhaps not easily generalizable to other situations, so a different approach is preferable.
Second solution (preferred).
For each element of $K' - F'$ that also occurs as an element $f \in F$, replace that copy of $f$ with the pair $(f,\alpha)$, where $\alpha$ is the smallest ordinal such that $(f, \alpha) \not\in K' \cup F$. Such an ordinal exists, because otherwise there would be an injection of the proper class of all ordinals into the set $K' \cup F$.
Third solution (with the axiom of choice).
It is enough to prove the existence of some set $A$ that has the same cardinality as $K' - F'$ which is disjoint from $F$. Then the field structure on $K'$ can be transported onto $A \cup F$ in the obvious way. Of course, the second solution accomplishes this, but it's really easy to see using cardinal arithmetic.
Namely, pick some infinite set $S$ that is strictly larger than both $K'$ and $F$. (For example, we could take $S = P(K' \cup F) \cup \mathbf{N}$.) Then using the axiom of choice, $S - F$ has the same cardinality as $S$, which itself contains a set equipotent to $K' - F'$.
This relies on the fact that if $B$ and $C$ are sets with cardinality $< \kappa$ (with $\kappa$ infinite), then $B \cup C$ also has cardinality $< \kappa$.
A: Do you have any problem considering that $\mathbb R\subset \mathbb C$? Because $\mathbb R$ is not a subset of $\mathbb C$.
What you are really doing is identifying (your word) the real number $a\in\mathbb R$ with the complex number $a + 0i \in\mathbb C$.
Nobody has a problem with that because the identification can be used to define the copy of $\mathbb R$ living inside $\mathbb C$ in such a way that we really just replace the copy with the original for practical purposes.
So in the end, there's no difference between (1) thinking of the larger space as an extension of the smaller, and (2) thinking of the larger space as containing a copy of the smaller.
