How to find all ideals of $A = \Bbb Z\times K,$ where $K$ is a field? This is the task: Find all ideals of $A = \Bbb Z\times K$ where $K$ is a field, with addition and multiplication defined by $$(m,x)+(n,y)=(m+n,x+y)$$ and $$(m,x)\cdot (n,y)=(mn, my+nx+xy).$$
I have no idea how to find the ideals. I know what an ideal is, but I have not found how can I accurately find all ideals (is there no algorithm or something like this?). 
 A: Hint: For any ideal $I \subset \Bbb Z$, you'll find that $\{(m,x): m \in I, x \in K\}$ will be an ideal of $\Bbb Z \times K$.  In fact, these will be the only ideals.

Edit: I missed something. If $K$ is a field of characteristic $p$, then $mp\Bbb Z \times \{0\}$ is also an ideal for any $m \in \Bbb N$.  I think that with that, we have now accounted for all ideals.

Explanation in other words:
For example, take $K = \Bbb Z_p$, the integers modulo $p$.  For every $m \in \Bbb Z$, the set $m\Bbb Z \times \Bbb Z_p$ will be an ideal of $A = \Bbb Z \times \Bbb Z_p$.  The set $pm\Bbb Z \times \{0\}$ will also be an ideal.  There are no other ideals in $A$.
A: For better clarity, I'll denote by $\def\1{\mathbf{1}}\1$ the identity of $K$.
The set
$$
I_{\mathbb{Z}}=\{m\in\mathbb{Z}:(m,x)\in I\text{, for some }x\in K\}
$$
is an ideal of $\mathbb{Z}$. Indeed, if $(m,x),(n,y)\in I$, then $(m+n,x+y)\in I$, so $m+n\in I_{\mathbb{Z}}$. If $(m,x)\in I$ and $n$ is an integer, then $(m,x)(n,0)=(mn,nx)\in I$ and $mn\in I_{\mathbb{Z}}$.
Therefore $I_{\mathbb{Z}}=a\mathbb{Z}$, for a unique $a\ge0$.
Let $(m,x)\in I$. If $m\1+x$ is invertible, then we can solve for $y$ in
$$
(m,x)(1,y)=(m,z)
$$
for every $z\in K$, because we can choose $y=(m\1+x)^{-1}(z-x)$. So, if for every $m\in I_{\mathbb{Z}}$ there exists $x\in K$ with $(m,x)\in I$ and $m\1+x$ invertible, then $I=I_{\mathbb{Z}}\times K$.
It remains to analyze the case when, for every $m\in I_{\mathbb{Z}}$ and every $x\in K$ such that $(m,x)\in I$, we have $m\1+x$ not invertible, which means $x=-m\1$.
Thus $I=\{(am,-am\1):m\in\mathbb{Z}\}$, where $I_{\mathbb{Z}}=a\mathbb{Z}$. Is this an ideal? It is certainly closed under sums and contains zero. For arbitrary $(n,y)$, we have
$$
(am,-am\1)(n,y)=(amn,amy-amn\1-amy)=(amn,-amn\1)
$$
