TITLE: Minimal cover v.s random reals
QUESTION [9 upvotes]: The following set theoretical question is inspired by a question from recursion theory:

Question: Is there an $L$-random real $r$ which is a minimal cover over another real $x$?

Where a minimal cover $r$ over $x$ means that $x\in L[r] \wedge r\not\in L[x]$ but there is not real $z$ so that $x\in L[z]\wedge z\in L[r]\wedge r\not\in L[z]\wedge z\not\in L[x]$.

REPLY [6 votes]: Suppose $r$ is random over $L$, $x \in 2^{\omega} \cap L[r]$ and $r \notin L[x]$. For $y \in 2^{\omega}$, let $y_0, y_1 \in 2^{\omega}$ denote the even and odd parts of $y$ - So $y = y_0 \oplus y_1$. Let $B$ denote the random algebra and $A$, the complete subalgebra of $B$ generated by $x$. Let $G_A$ be an $A$-generic filter over $L$ such that $L[x] = L[G_A]$. Then $L[r]$ is obtained by adding a $B/G_A$-generic filter over $L[x]$. Since this is same as adding a random real over $L[x]$, we can choose a real $y \in L[r]$ such that $y$ is random over $L[x]$ and $L[x][y] = L[r]$. Let $z = x \oplus y_0$. Then $L[x] \subsetneq L[z] \subsetneq L[r]$.