Minimum transitive models and V=L Is there a c.e. theory $T⊢\text{ZFC}$ in the language of set theory such that the minimum transitive model of $T$ exists but does not satisfy $V=L$?
You may assume that ZFC has transitive models.  Note that $M$ is minimal iff $∀M' \, M'⊈M$ and minimum iff $∀M' \, M'⊇M$.
It may be tempting to consider ZFC + $0^\#$ (assuming large cardinal axioms), but while this theory has a minimum inner model (i.e. $L[0^\#]$), it has incomparable minimal transitive models.  Model comparability uses iterability, but transitiveness does not suffice for iterability.  Moreover, for every c.e. $T⊢\text{ZFC\P}+0^\#$ having a model $M$ with $On^M = α < ω_1$, the intersection of all such $M$ equals $L_α$, and furthermore a subset of $L_α$ is definable (with parameters) in all such $M$ iff it is in $L_{α^{+,\mathrm{CK}}}$.  To see this (briefly), $0^\#$ allows $M$ to 'continue' $L$ beyond $α$, and $L_{α^{+,\mathrm{CK}}}⊆(L_{α^{+,\mathrm{CK}}})^M$ (the well-founded part of any model of KP being admissible), and so $L_{α^{+,\mathrm{CK}}}∩V_α=L_α$.  Also, existence of $M$ is $Σ^1_1(α)$, so the intersection of all $M$ is at most $L_{α^{+,\mathrm{CK}}}$.
A minimum transitive model of ZFC + $A$ for a statement $A$ cannot be produced through set forcing either.  Also, I think that minimal models need not satisfy $V=HOD$; not sure about minimum models.
Thus, minimum transitive models usually either satisfy $V=L$ or do not exist.  However, I suspect that exceptions exist, including in the form $L_α[r]$ with $r∈ℝ$, but require an interesting coding argument.  Perhaps there is such an $r$ computable from the theory of the least transitive model of ZFC $L_α$, with $r$ self-verifying (in $L_α[r]$) and yet weak enough for $L_α[r]⊨\text{ZFC}$.
Update:  I accepted Farmer Schlutzenberg's positive answer (see also the partial answers by Hamkins and Enayat).  A remaining open problem is whether such a $T$ can be obtained by extending ZFC with a single statement.
 A: Yes, I claim you can in fact get one whose minimum model is a set forcing extension of a segment of $L$. Let $L_\alpha$ be least modelling ZFC. Let $\mathbb{P}=\mathbb{P}^{L_\alpha}$ be Jensen's forcing for adding a $\Pi^1_2$-singleton, as defined over $L_\alpha$; ZFC proves that forcing with $\mathbb{P}^L$ over $L$ adds a unique $L$-generic filter. Working in $V$, define the sequence $\left<p_n\right>_{n<\omega}$ through $\mathbb{P}$, which will be $L_\alpha$-generic, as follows. Fix a recursive enumeration $\left<\varphi_k\right>_{k<\omega}$ of all formulas in the language of set theory in one free variable. Now let $p_0=\emptyset$. Suppose we have defined $p_n$. Let $\alpha_n$ be the least $\beta$ such that $L_\beta\preccurlyeq_{\Sigma_n}L_\alpha$. Let $p_{n+1}$ be the least $p\in\mathbb{P}$ such that $p\leq p_n$ and $p$ is in all open dense subsets $D$ of $\mathbb{P}$ which are defined over some $L_{\alpha_i}$ by $\varphi_j$, for some $(i,j)$ with $i,j\leq n$. (That is, for some such $(i,j)$, $D$ is the unique $D'\in L_{\alpha_i}$ such that $L_{\alpha_i}\models \varphi_j(D')$.) This $p$ exists since there are only finitely many dense sets we consider here. This defines the sequence.
Let $G$ be the filter generated by $\left<p_n\right>_{n<\omega}$.
Since $L_\alpha$ is pointwise definable, $G$ is $L_\alpha$-generic. So $L_\alpha[G]\models$ ZFC.
Now let $T$ be the following recursive (not just r.e.) theory extending ZFC, which will correspond to the construction above: for $n<\omega$ let $T_n$ be ZFC + "$V=L[g]$ for some $(L,\mathbb{Q})$-generic filter $g$, where $\mathbb{Q}$ is Jensen's forcing in $L$, and defining $q_n$ with respect to $L$ as in the construction above, we have $q_n\in g$". (That is, note that there is a recursive sequence $\left<\psi_n\right>_{n<\omega}$ of formulas such that for each $n$, we have that $p_n$ is the unique $p\in L_\alpha$ such that $L_\alpha\models\psi_n(p)$. Have $T_n$ assert that there is $q\in g$ (where $g$ is as above) such that $L\models\psi_n(q)$. Thus, the sequence $\left<T_n\right>_{n<\omega}$ is recursive. Of course, the complexity of $\psi_n$ increases with $n$. Note that the "$g$" is a bound variable, not some new constant, so I am working with only the language of set theory.) Now set $T=\bigcup_{n<\omega}T_n$. So $T$ is recursive.
Claim: $L_\alpha[G]$ is the minimum transitive model of $T$.
Proof: Let $P$ be any transitive model of $T$. Then certainly $L_\alpha\subseteq P$, and easily if $\alpha<\mathrm{OR}^P$ then $L_\alpha[G]\subseteq P$. So we may assume $\alpha=\mathrm{OR}^P$. So $L^P=L_\alpha$. But then letting $G'$ be the (unique) $(L_\alpha,\mathbb{P})$-generic filter such that $P=L_\alpha[G']$, we get $p_n\in G'$ for each $n<\omega$, since $P\models T_n$ for each $n$. But therefore $G'=G$, so $P=L_\alpha[G]$, which suffices.
Remark: This construction is somewhat related to that for Proposition 34 of "On a Conjecture Regarding the Mouse Order for Weasels", arXiv:2207.06136, joint with Jan Kruschewski; (that proposition is stated rather generally, but in its simplest instantiation it gives an example of $G$ which is (only just) Cohen generic over $L_{\omega_1^{\mathrm{ck}}}$ but with KP failing in $L_{\omega_1^{\mathrm{ck}}}[G]$).
Remark 2: The question is rather related to a question of Harvey Friedman's, on which Woodin and Koellner made recent (boldface) progress. The question was (if I recall it precisely) whether there can be an ordinal $\alpha$ and a single sentence $\varphi$ such that there is a unique transitive model $M$ such that $\mathrm{OR}^M=\alpha$ and $M\models$ ZFC + "$V\neq L$" + $\varphi$. It was already known that any such model must satisfy "$0^\sharp$ does not exist", and I think also that it must satisfy "$V=\mathrm{HOD}$".
A: Here is another partial result; it complements Joel Hamkins' answer. Note that in the following theorem, $T$ is not necessarily a c.e. theory.
Theorem. Suppose $T$ is an extension of $\mathrm{ZF} + \exists a~\mathrm{V}=\mathrm{L}[a]$ that has a minimum transitive model $M$. Then $M$ satisfies $\mathrm{V = HOD}$.
Proof outline. By a classical result of Vopěnka there is a
partial order $\mathbb{P}$ in $\mathrm{HOD}^{M}$ such that the model $M$ is a $\mathbb{P}$-generic extension of $\mathrm{HOD}^{M}$.
Moreover, as shown by Grigorieff (see Theorem 1 of Sec. 5 of this paper
) $%
\mathbb{P}$ can be arranged to be weakly homogeneous.
Now let $G_1$ and $G_2$ be mutually generic $
\mathbb{P}$-filters over $\mathrm{HOD}^{M}$. For $i=1,2$ let $N_i$ denote $\mathrm{HOD}^{M}[G_i]$. By weak homogeneity of $\mathbb{P}$ we have:
$$(1)~~\mathrm{Th}(M) = \mathrm{Th}(N_1)= \mathrm{Th}(N_2).$$
On the other hand, by an old argument of Solovay, the mutual genericity of $G_1$ and $G_2$ over $\mathrm{HOD}^{M}$ implies:
$$(2)~~N_1 \cap N_2 = \mathrm{HOD}^{M}.$$
(1) and (2) together contradict the assumption that $M$ is a minimum model of $T$, thus $M$ satisfies $\mathrm{V = HOD}$.
A: This is not a full answer, but I found it interesting to notice that if we relax the c.e. requirement somewhat, then there is a sweeping positive answer.
Theorem. Every complete theory extending ZFC + V=HOD has a minimum transitive model, if it has any transitive models.
Proof. Suppose $T$ is a complete theory extending ZFC + V=HOD. Since there is a definable global well order in this theory, we have definable Skolem functions. Therefore, in any model of $T$ the parameter-free definable objects will form an elementary substructure. This model will be pointwise definable, and a copy of it will be contained in all other models of the theory. So it will be a minimum model of $T$. $\Box$
The essence of the idea is that the pointwise definable models of ZFC are exactly the prime models of the theory ZFC + V=HOD.
Corollary. Every transitive model of ZFC + V$\neq$L + V=HOD sits above a minimum model of its theory.
So this provides instances of your requested phenomenon using $T=\text{Th}(M)$, where $M$ is any transitive model of ZFC + V$\neq$L + V=HOD.
Meanwhile, similar thinking leads to a negative answer for arithmetically definable complete theories.
Corollary. If a complete theory $T$ extends ZFC +V=HOD and is arithmetically definable, then it has no transitive models.
Proof. If it had a transitive model, then it would have a minimum transitive model $M$, which must be pointwise definable. But the theory $T$ is arithmetically definable, and would therefore be an element of $M$. But then $M$ would be able to define a copy of itself inside itself by consulting the theory — the model $M$ is uniquely isomorphic to the set of definable elements inside any model of $T$. But this is impossible since $M$ thinks that uncountable sets exist. $\Box$
One can relax arithmetically definable to hyperarithmetic — one just wants to know that the theory must be inside any transitive model of the theory.
