Probably a really easy question...does the following ring exist? This may be an easy couplet of questions but I have never seen a ring of this form ever brought up:

*

*Does the following exist as a ring:

$R=\mathbb{Z}[x]/\{x^2=0; 7x = 0\}$
2.Does $R$ have the form $R=\{ax+b; b \in \mathbb{Z}; a \in \mathbb{F}_7\}$,
where $a'x+b'$ and $ax+b$ are distinct elements as long as either $a' \not = a$ or $b'\not = b$. Put another way, for example, $6x$ and $13x$ are the same element in $R$, but $6$ and $13$ are distinct elements in $R$.
If the answer to either is NO, then why not.
Thanks for looking at this!
 A: Yes, $R$ certainly exists as a ring.  If nothing else, $R=\Bbb Z[x]/(x^2, 7x)$.  The question is whether that ring is isomorphic to a more familiar ring.  The answer is that I don't think so.
It's not $\Bbb F_7 \oplus \Bbb Z$ as suggested in another answer (now deleted) because all products of two elements of the first component are $0$ and because multiplication isn't performed componentwise.  I guess you could define a trivial ring structure on the additive group $\Bbb Z/7 \Bbb Z$, give that a name, and define an appropriate product across components, but it seems easier just to work with the quotient ring.
All elements of $R$ can be presented in the form you suggest.  Just be careful to remember, when multiplying, that $x^2=0$.
A: Starting with an abelian group $G$, there is a construction called the trivial extension $\mathbb Z\ltimes G$ which creates a ring whose underlying abelian group is $\mathbb Z\times G$ and whose multiplication is defined by
$(n, a)(m, b)=(nm, n\cdot b+m\cdot a)$
This is isomorphic to your ring via the map $R\to \mathbb Z\ltimes C_7$ given by
$\sum_{i=0}^k n_ix^i + I\mapsto (n_0, n_1\cdot c)$
Here I have used $I$ to denote your ideal $(x^2, 7x)$, and I have chosen $c$ to be a generator of the additively written cyclic abelian group $C_7$.
