How would the constructible hierarchy change if you remove parameters? The constructible hierarchy is defined as follows.  $L_0=\varnothing$.  For any ordinal $\beta$, $L_{\beta+1}(X)=Def(L_{\beta})$, where $Def(X)$ is the set of all subsets of $X$ which are first-order definable using elements of $X$ as parameters.  For any limit ordinal $\gamma$, $L_\gamma=\cup_{\beta<\gamma}L_\beta$. Finally, $L=\cup_\alpha L_\alpha$.
My question is, what if instead of taking the set of all subsets definable using parameters, we take the set of all subsets definable without parameters?  How much would the constructible universe shrink?
 A: This question shows up as Exercise 23 at the end of Chapter VI in the old version of Kunen's textbook. The claim in the textbook is that the structure defined in the way suggested by the OP is just $L$. I do not know how to show this, but I'm providing a partial answer here to bump this question up.
More specifically, I claim the following: for any $\alpha$ and any $x\in L_{\alpha+1}$, $x$ is definable over $L_\alpha$ with only parameters of the form $L_\xi$, for some $\xi<\alpha$.
We will show this by induction on $\alpha$. Suppose this claim holds below $\alpha$, and suppose
$$
x=\{z\in L_\alpha\mid L_\alpha\vDash \varphi(z, p_1,...,p_n)\}
$$
Then we may fix some $\gamma<\alpha$ with $p_1,...,p_n\in L_{\gamma+1}$. But the claim holds for $L_\gamma$, which means the $p_i$'s can be defined over $L_\gamma$ with only parameters of the form $L_\xi$. For each $p_i$, fix a formula $\psi_i(z)$ (with the appropriate parameters plugged in) that defines $p_i$. And let $\psi_i'(z)$ be the modification of $\psi_i(z)$ by bounding all unbounded quantifiers with $L_\gamma$.
Now it follows that each $p_i$ is definable over $L_\alpha$ in the following way:
$$
p_i=\{z\in L_\alpha\mid L_\alpha\vDash (z\in L_\gamma)\wedge \psi_i'(z)\}
$$
And our set $x$ can be defined as follows:
$$
x=\{z\in L_\alpha\mid L_\alpha\vDash \exists p_1...\exists p_n\varphi(z,p_1,...,p_n)\wedge \\(\forall v)(v\in p_1)\leftrightarrow v\in L_\gamma\wedge \psi_1'(v) \wedge...\wedge\\
(\forall v)(v\in p_n)\leftrightarrow v\in L_\gamma\wedge \psi_n'(v) \}
$$
Basically, the idea is that if your parameters can be defined in this restricted way, then so can you.
A: Let $L' = \bigcup_\alpha L'_\alpha$ be the other class. We shall use induction on $x\in L$ via the ordering the $<_L$ to show $x\in L'$. Now supposing $y\in L'$ for $y<_Lx$ and that $x$ is defined over $L_\alpha$ via some formula $\varphi(u, v_1,\dots, v_n)$ and parameters $y_1,\dots, y_n$. Define $y_{n+1} = L_\alpha$ for notational convenience and also suppose $x\not = L_\alpha$ and so $L_\alpha<_L x$, we shall deal with the equality case separately. Now since for each $i\le n+1$, $y_i <_L x$, by the induction hypothesis we must have an ordinal $\beta_i$ and a formula $\psi_i(u)$ such that $$y_i = \{ u \in L'_{\beta_i}: L'_{\beta_i}\models \psi_i(u)\} \in L'_{\beta_i+1}.$$ Also let $G:ORD^{n+1}\rightarrow ORD$ be the canonical Gödel pairing function. Now let $\beta = G(\beta_1, \dots, \beta_{n+1})$. So let $\beta = \xi + m$ where $\xi$ is a limit ordinal and $m$ is a natural number. We shall define $x$ over $L'_{\xi+\omega}$, so $x\in L'_{\xi+\omega+1}$. We can see that $$x=\{u \in L'_{\xi+\omega}: L'_{\xi+\omega}\models "\exists \eta(\eta \text{ is the largest limit ordinal and if you let } \langle \theta_1,\dots, \theta_{n+1}\rangle = G^{-1}(\eta+m), \text{ and if } z_1, \dots, z_{n+1} \text{ are defined like: } z_i \text{ is the element of } L'_{\theta_i+1} \text{ defined by } \psi_i(v) \text{ over }L'_{\theta_i}, \text{ then } u\in z_{n+1} \text{ and } z_{n+1}\models \varphi(u, z_1,\dots, z_n)"\}$$
So we coded our parameters in $\beta$ and recovered them in the formula. So in the formula $\eta$ is going to be $\xi$ and $\theta_i =\beta_i$ and $z_i=y_i$. And the only parameter we used is $m$, which is a natural number, and so since it is a fixed finite object, we can hard code it into our formula. Also the fact that the definition works needs a bit of checking, but it follows from the fact that $L'_{\xi+\omega}$ is transitive. Also you can easily see that if $\zeta$ is a limit ordinal then $L'_\zeta\cap ORD = \zeta$, which is also used to check the details.
For the case $x=L_\alpha$ note that a similar argument as above works. Just take $\beta$ large enough such that $L_\alpha\subset L'_\beta$. Then look at $L'_{\beta+\omega}$. Then take the definable set over it defined by $\varphi(u) = \exists \xi u \in L_\xi$, call it $y$. Then since $L'_{\beta+\omega}$ is transitive either $y=L_\alpha$ or $L_\alpha\in y$, which solves this case.
I apologize if this is a bit sketchy, but the thing is that writing it out fully and formally would make it very long. Please do tell me if you spot any errors.

Edit: my argument for the case $x=L_\alpha$ is not completely correct, since it fails for successor $\alpha$. To fix this, you may assume we were working with the $J$-hierarchy indexed with limit ordinals all along and consider the formula: $\varphi(u) = \exists\xi u \in S_\xi$.
