Can a standard model of $\sf ZFC$ contain all the ordinals without being transitive?

Question

By a standard model of ZFC I mean a model of ZFC that can be a set or proper class and whose elementhood relation is the true elementhood relation. A transitive model of ZFC is a standard model of ZFC that is also a transitive class. By the Mostowski collapse lemma, every standard model of ZFC is isomorphic, via a unique isomorphism, to a unique transitive model of ZFC. My question is, can one always "uncollapse" a transitive model of ZFC to a standard but non-transitive model of ZFC? For instance, is there a non-transitive standard model of ZFC isomorphic to the minimal inner model $L$? And also, is there a non-transitive standard model of ZFC containing all the ordinals that is isomorphic to the minimal inner model $L$?

Edit: Given the helpful comments made (the answer to the first two questions is yes), the only question I have remaining is the last: Is there a non-transitive standard model of ZFC containing all the ordinals that is isomorphic to the minimal inner model $L$? If not, then $L$ is not only the minimal inner model, it is the minimal standard model containing all the ordinals. More generally, is there a non-transitive standard model of ZFC containing all the ordinals?

Answer 1

I've deleted my previous answer since it was wrong, as pointed out by Rodrigo Freire in the comments.

In fact, it is possible to have a non-transitive model whose ordinals are an initial segment of the ordinals. Say that $M$ is a transitive model such that $M\neq V_\alpha$ for any $\alpha\in\rm Ord\cup\{Ord\}$ (where $V_{\rm Ord}$ is just the whole universe). Then there is a smallest $\alpha$ such $\alpha\in M$ and $\mathcal P(\alpha)^M\neq\mathcal P(\alpha)$.

Define $N$ to be the model obtained by recursively replacing $\mathcal P(\alpha)^M$ by $\mathcal P(\alpha)$, or even just adding one new set to this collection. Then $N$ is a standard model, its ordinals are an initial segment of the ordinals, but it is not transitive.

If we take $M=L$ and $V\neq L$, then we can of course obtain a model of $V=L$ which is not $L$.

Answer 2

Asaf Karagila has answered the question, but I have been thinking about partial results for the minimality of $L$ in the direction of his previous answer, as asked by Jesse Elliot in his final paragraph.

First, excuse-me for saying that I think set theory has not used standard models (in the sense of this question) very much because they are isomorphic to transitive models. So, we are not very used to them. However, in fact it is easy to "uncollapse" a transitive model $M$: take an element $a\in M$ and replace it everywhere transitively by $a\cup \left\{a\right\}$. If $a$ is not an ordinal, then the resulting standard model will share the ordinals of $M$.

Now, on a more positive direction, let us investigate a partial minimality result for $L$:

-Let $M\subseteq L$ be a standard model such that its ordinals are the real ordinals. Then $M=L$ iff the constructible order $Od$ (see Shoenfield, ML, page 272) is absolute for $L^M$.

proof: First notice that $L^M=\left\{x\in M : (x\in L)^M\right\}$ is a standard model whose ordinals are the real ordinals. If $L^M$ were transitive, then it would include $L$, hence $M$ would be equal to $L$. So, let us assume that $L^M$ is not transitive.

Let $K$ be the transitive collapse of $L^M$. The image of $K$ is a transitive model of $ZF$ containing all ordinals and contained in $L$, so it is $L$. Let $x$ be a minimal counterexample to the transitivity of $L^M$. Then $K(x)\neq x$, so $Od(K(x))\neq Od(x)$ (recall that $M\subseteq L$, hence $Od$ is defined for all elements of $M$ and is injective). Since $K$ is an isomorphism from $L^M$ to $L$, $K(Od^{L^M}(x))=Od(K(x))$. From the absoluteness hypothesis, $Od^{L^M}(x)=Od(x)$.

Therefore,

$K(Od(x))=K(Od^{L^M}(x))=Od(K(x))\neq Od(x)$,

so $Od(x)$ is an ordinal which is moved by $K$. This is a contradiction with the hypothesis that the ordinals of $M$ are exactly the ordinals.