Given a forcing extension $V^F$ of a set-theoretic universe $V$, how much sense does it make to talk about preservation of algebraic structures? For example, if one has a ring quotient $Q=R/I$, does it make sense to write $Q^F=R^F/I^F$?
For example, suppose we have a hyperreal field $Q=R/I$ where $R$ is the ring of sequences of real numbers. What does this look like in $V^F$? Which part remains the same and which is reinterpreted?
The kind of application I have in mind is detailed here.
On
Let me add a bit to Asaf's answer, re: versions of the hyperreals.
Let's look at bog-standard ultrapowers first.
First of all, a quick observation: if $\mathcal{U}$ is an ultrafilter in $\mathbb{N}$, and in $V[G]$ $\mathcal{U}'$ is an ultrafilter on $\mathbb{N}$ extending $\mathcal{U}$, then there is a natural embedding of $(\mathbb{R}^\mathbb{N})^V/\mathcal{U}$ into $(\mathbb{R}^\mathbb{N})^{V[G]}/\mathcal{U}'$: if I have two sequences in $V$ which are $\mathcal{U}$-inequivalent, then they're also $\mathcal{U}'$-inequivalent since $\mathcal{U}\subseteq\mathcal{U}'$. (I've used $\mathbb{N}$ as an index set since it has a forcing-absolute definition; if we change the index set, then we again run into the "take the set vs. move the definition" issue.) Moreover, we can show that this is in fact an elementary embedding (via Tarski-Vaught + Los' theorem).
A natural question at this point is when $\mathcal{U}'$ can be canonically chosen. This is a bit vague, but one natural version is to request that it be unique: that is, that $\mathcal{U}$ extends to a unique ultrafilter in $V[G]$. This is a very stringent condition; Joel David Hamkins gave a combinatorial characterization of the forcings with this property, and I don't know offhand of any such interesting forcings. EDIT: See Vladimir Kanovei's comment below.
On the other hand, we usually don't pick a single hyperreal field in the first place - for the vast majority of the time (so far as I'm aware), we just need a (sufficiently saturated, perhaps) hyperreal field. So having many possible extensions isn't necessarily bad, especially since the canonical embeddings are elementary.
A better approach might be via Kanovei-Shelah. They produce a formula $\varphi$ in the language of set theory which ZFC-provably defines a nonstandard model of the theory of the real numbers. This formula can therefore be interpreted in $V[G]$ as well as in $V$. The task, then, is to compare $\varphi^V$ with $\varphi^{V[G]}$ - in particular, in $V[G]$ there should be a canonical elementary embedding $j: \varphi^V\rightarrow \varphi^{V[G]}$. (Since $V[G]\supseteq V$, there's no problem talking about $\varphi^V$ inside $V[G]$: in fact, a lovely result due independently to Laver and Woodin states that $V$ will always be definable in $V[G]$! I think that's absolutely wild.)
Unfortunately, I'm not familiar enough with the K-S construction to be confident in the existence of $j$. I'd be incredibly surprised if such a thing didn't exist, especially since the K-S model is built by amalgamating ultrapowers in a certain way, but their construction is complicated enough that a quick glance isn't enough to resolve the issue for me.
What if we still want to work with ultrafilters, but not be restrictive on our forcing notions and also not make arbitrary choices? Well in one sense we're out of luck; but in another sense, there's still something we can do. Namely, any filter $\mathcal{F}$ on an index set $I$ will induce the reduced power $(\mathbb{R}^I)/\mathcal{F}$; and if $\mathcal{F}\subseteq\mathcal{F}'$, we have a canonical surjective homomorphism $$h_{\mathcal{F}\subseteq\mathcal{F}'}:(\mathbb{R}^I)/\mathcal{F}\rightarrow(\mathbb{R}^I)/\mathcal{F}'.$$ These homomorphisms compose in the desired way: $h_{\mathcal{F}\subseteq\mathcal{F}'}\circ h_{\mathcal{F}'\subseteq\mathcal{F}''}=h_{\mathcal{F}\subseteq\mathcal{F}''}$.
This turns the set of reduced powers (over a fixed index set) into a poset $\mathbb{P}_I$ - in fact, a lower semilattice - in which the genuine ultrapowers are exactly the maximal elements. (We can get a genuine lattice $\mathbb{L}_I$ if we allow the whole powerset to be a filter; the corresponding reduced product is then the one-element structure. Here the genuine ultrapowers are the coatoms. Note that $\mathbb{P}_I$ and $\mathbb{L}_I$ differ by one element: $\mathbb{L}_I$ is gotten from $\mathbb{P}_I$ by adding a "top.")
So what? Well, each filter $\mathcal{F}\in V$ extends to a canonical filter $\hat{\mathcal{F}}\in V[G]$. Of course, $\hat{\cdot}$ need not preserve ultra-ness, but that's not a problem for us. So in $V[G]$ we get a canonical embedding $c:(\mathbb{P}_I)^V\rightarrow(\mathbb{P}_I)^{V[G]}$. (And identically a canonical embedding $d:(\mathbb{L}_I)^V\rightarrow(\mathbb{L}_I)^{V[G]}$.)
So, to the extent to which we can replace reasoning in a single hyperreal field with reasoning in the entire "spectrum of reduced products," we get complete canonicity quickly and easily. The good news is that it's pretty automatic to do this; the bad news is that it's pretty clunky as well. But I actually prefer the structures $\mathbb{P}_I$ and $\mathbb{L}_I$ to any particular ultrapower, so for me this is a nice thing to do.
You need to discern between an algebraic structure (or any first order structure) and its definition.
Since first order logic is absolute, structures are always preserved. You don't add new elements to ground model structures.
However, definitions can change. Things like the Baire space, or the hyperreal fields, can easily change since new sequences can be added, as can new ultrafilters.
If you talk about the concrete sets, then they do not change, and marking them with the superscript of the universe you work in is meaningless. But if you think about your structure as a definition being reinterpreted in a new model, then it makes sense.
To your question, the hyperreal numbers form a tricky part here.
For the one part, if $U$ was an ultrafilter on $\Bbb N$ in $V$, then in $V^F$ it might not even be a filter anymore, but even the filter it generates might not be an ultrafilter anymore.
This means that even though that $(\Bbb{R^N}/U)^V$ is a field, both $\Bbb R$ and $\Bbb{R^N}$ have significantly changed, and $U$ is not even an ultrafilter anymore, nor it generates one (or rather it is possible that it no longer generates one). So while $\Bbb R^V$ is a subfield of $\Bbb R^{V^F}$, it is not complete, and $(\Bbb R^V)^\Bbb N/U$ is not even a field, as this is just a reduced-product rather than an ultrapower.
I guess, although I haven't checked it, that if $U^*$ is an ultrafilter extending $U$ in $V^F$, then $(\Bbb{R^N}/U)^V$ has some natural embedding into $\Bbb{R^N}/U^*$ in $V^F$. But I'd imagine that the exact nature of this embedding might depend a lot on the type of forcing, and the properties of $U$ and $U^*$ (its second and third order properties more than its first-order ones, anyway).