Atoms necessary for the existence of a generic filter?

I'm reading Some applications of the method of forcing by Todorchevich and Farah and one of the exercises seems to be wrong to me.

Definitions

We fix some partially ordered set $(P,\preceq)$.

A filter on the poset is a set $G\subseteq P$ such that

1. whenever $p\in G$ and $p\preceq q$ then $q\in G$, and
2. for all $p,q\in G$, there is $r\in G$ such that $r\preceq q$ and $r\preceq p$.

The set $D\subseteq P$ is dense if for all $p\in P$, there is $q\in D$ such that $q\preceq p$.

An atom is an element $p\in P$ such that the set $\{q\in P:q\preceq p\}$ is linearly ordered by $\preceq$.

Exercise

There is a filter $G$ that intersects every dense set (a generic filter) if and only if there is an atom in $P$.

Counterexample

Let $P=\mathbb{Z}$ and let $p\preceq q$ iff $|p|>|q|$ or $p=q$. Then $(P,\preceq)$ has no atom, but $G=P$ is a filter that intersects every subset and hence every dense subset.

Is there something wrong with my counterexample? If not, is there some version of the exercise that is related to forcing and is correct?

Since there was some speculation what is actually in the text, here is the relevant part of the first page:

Solution 1:

The authors are probably assuming that for every $p$ there are $q,r < p$ with $q \perp r$. This is a consequence of a property of the poset known as separativity.

A (nonempty) poset $P$ that is separative has no generic filter. To see this, let $G$ be a generic filter on $P$ and let $p \in P$. Pick $q,r < p$ as above with $q \perp r$. Then at most one of $q$ and $r$ can be in $G$, because they are incompatible. In particular $G \not = P$. Moreover, we see that for every $p \in P$ there is some $s < p $ with $s \not \in G$. Thus the set $P \setminus G$ is dense, and $G$ cannot possibly meet this set. Hence there is no generic filter on $P$.

It appears to me that when the authors defined an atom $p$, they should have said that all elements below $p$ are compatible, not that they are linearly ordered. If $p$ is an atom in this sense, the filter consisting of the upward closure of the set of elements less than $p$ will be generic, and the above argument shows that if there is no atom in this sense then there is no generic filter. The actual content of the text (posted in the question) doesn't appear to be correct, because if $P$ itself is a filter then it will be a generic filter, regardless whether any elements of the poset generate linearly ordered ideals.