Yes, All Fields are rings, and all rings are groups. That said, it is perhaps worthwhile to add a few words of clarification.
Among these three, fields, rings and groups, the groups have the simpler structure. Groups require only one operation among it's members, and it is this operation that needs to satisfy the group axioms. Rings, however, require two operations, sometimes called "addition" and "multiplication", but this is only a convention. These two operations have to cooperate with each other -- this is actually one of the ring axioms. You want to have something like the distributive law that connects addition and multiplication of numbers. This is possibly the reason why the two operations of a ring are often called "addition" and "multiplication".
What might be confusing for beginners, is that with respect to one of these operations - traditionally the "addition", the ring-structure is required to be a group, while the second operation - the "multiplication", it is not required to induce a group, and indeed, one of the most important rings -- the ring of real valued $n\times n$ matrices, has two operations -- addition and multiplication -- and while with respect to the addition operations the matrices form a group, with respect to the multiplication operation they (after we've discarded the zero element) do not form a group, because some matrices do not have a multiplicative inverse. The ring of matrices has a multiplicative identity element, but this is not required by definition. There are rings without a multiplicative identity.
So a ring is a group with an additional structure - that obtained by another operation. A field is a very special kind of ring. Not only do we require that we have two groups with respect to both operations, we also require them to be commutative. This is quite a restriction compared to rings. More precisely, a field is a triple $\{F,+\cdot\}$ such that
$\{F,+\}$ is a commutative group and $\{F^*,\cdot\}$ is also a commutative group, where $F^*$ is obtained from $F$ by discarding the additive identity (normally denoted by $0$), such that the operations $+$ and $\cdot$ satisfy the distributive law.