TITLE: Morphisms of $\mathbb E_l$-rings between $\mathbb E_k$-rings for $l<k$
QUESTION [8 upvotes]: Given two commutative rings $A$ and $B$, any map of rings $A\to B$ will automatically preserve the commutative structure. This is to say, the forgetful functor $\operatorname{CRing}\to \operatorname{Ring}$ is fully faithful.
The analogous situation in brave new algebra is richer: we have $\mathbb E_k$-rings for all $1\le k \le \infty$, encompassing the notion of commutativity for up to $k$ factors.
Q: Does a result such as above still hold, i.e. is the forgetful functor $\operatorname{Alg}_{\mathbb E_k}\to \operatorname{Alg}_{\mathbb E_l}$ for some fixed $1\le l < k\le \infty$ fully faithful?
I would not be surprised if the answer turned out to be negative. After all, an $\mathbb E_k$-structure is not a condition but genuine structure instead, so a functor preserving it seems like a stronger condition. If that is the case, I would be very grateful if some examples could be provided of $\mathbb E_k$-ring maps that are not $\mathbb E_{k+1}$-ring maps (or just not $\mathbb E_\infty$-ring maps).
Of course, if the answer is negative, it would seem natural to ask under what conditions on an $\mathbb E_k$-rings $A$ and $B$ does every $\mathbb E_l$-ring map $A\to B$, for some fixed $l < k,$ give rise to an $\mathbb E_k$-ring map.

REPLY [9 votes]: Let's consider these in characteristic zero, so that we can use differential graded algebras and so that we can model $E_\infty$ things by strictly commutative things and $E_1$ things by associative things.
Let $A$ be the commutative DGA $\Bbb Q[x,y]$ where $x$ and $y$ are in degrees $2n$ and $2m$ respectively, with $dx = dy = 0$. Then the set of maps $A \to B$ in the homotopy category of commutative DGAs can be naturally identified with $H_{2n} B \times H_{2m} B$.
Now let's think of $A$ as an associative DGA; it's no longer semifree, so to compute maps in the homotopy category we first have to find a resolution $\tilde A \to A$. Here's one: the associative $DGA$ given by 
$$
\tilde A = \Bbb Q\langle x,y,z \rangle
$$
where $z$ is in degree $2n+2m=1$ and $dz = xy - yx$. Then $\tilde A$ is semifree, and there is a natural projection map $\tilde A \to A$ that sends $z$ to zero. With some work, one finds that it is a quasi-isomorphism.
As a result, the set of maps $A \to B$ in the homotopy category of $E_1$-algebras are the same as maps $\tilde A \to B$ in the homotopy category of $E_1$-algebras.
Let $B = \Bbb Q[z] / (z^2)$ with $|z| = 2n + 2m + 1$ and $dz = 0$. Then there is only the trivial map $A \to B$ sending $x$ and $y$ to zero, but the projection $\tilde A \to B$ sending $x$, $y$, and $z^2$ to zero is a nontrivial map in the homotopy category. This shows that the forgetful map is not full.