Base change of schemes preserves $\mathbb{A}^1$-contractibility

Question

Suppose $L/k$ is a field extension and $X \in Sm/k$ is an $\mathbb{A}^1$-contractible scheme over $k$. Then is the base change of $X$ i.e. $X_L = X \times_{Spec \ k} Spec \ L \in Sm/L$ is $\mathbb{A}^1$-contractible? Recall that a scheme $X \in Sm/k$ is called $\mathbb{A}^1$-contractible if $X \cong Spec \ k$ in Morel-Voevodosky $\mathbb{A}^1$-homotopy category $\mathcal{H}(k)$ and $Sm/k$ is the category of smooth seperated finite type schemes over $k$ endowed with the Nisnevich topology. May be we can assume here seperable extension and characteristic zero as a particular case.
Any reference or suggestion regarding this question is highly appreciated.

Answer 1

Yes.

(EDIT : Brian Shin pointed out to me that I absolutely did not need smoothness for that argument - I got confused in thinking it was needed to define $Sm/k \to Sm/L$ but of course a pullback of a smooth scheme is smooth)

$\mathcal H(k)$ is (the homotopy category of) the Bousfield localization of space-valued (simplicial) sheaves on $Sm/k$ at the maps $X\times \mathbb A^1\to X$ for all $X \in Sm/k$.

In particular, the basechange $Sm/k\to Sm/L$ induces a functor $Sh(Sm/k)\to Sh(Sm/L)$ on sheaf categories (given by left Kan extension followed by sheafification), which sends representables (so smooth schemes $X/k$) to their basechange (because the representable functor $F = \hom(-,X)$ is mapped to $\hom(-,X_L)$, which is already a sheaf, because the Nisnevich topology is subcanonical), and so it send $X\times \mathbb A^1\to X$ to $X_L\times\mathbb A^1\to X_L$, in particular we get a commutative square of localizations :

$$\require{AMScd}\begin{CD}Sh(Sm/k) @>>> Sh(Sm/L) \\ @VVV @VVV \\ \mathcal H(k) @>>> \mathcal H(L) \end{CD}$$

So if "down" sends $X\to *$ to an equivalence, so does "down-right", and therefore so does "right-down", but "right" sends it to $X_L\to *$, which proves the claim.