A certain valuation of $k(X,Y)$ with value group $\mathbb{Z}+\mathbb{Z}\alpha$ 
Let $k$ be a field, $X$ and $Y$ indeterminates, and suppose that $\alpha$ is a positive irrational number. Then the map $\nu:k[X,Y]\rightarrow \mathbb{R}\cup \{\infty\}$ defined by $\nu\left(\sum c_{n,m}X^nY^m\right) = \min\{n+m\alpha\}$ determines a valuation of $k(X,Y)$ with the value group $\mathbb{Z}+\mathbb{Z}\alpha$.

To be honest, I don't even know where to start with this problem. I've been lost for weeks. This isn't homework, it's just an exercise in the chapter of the book we just finished. Please include the definition of a valuation in your answer, as my book uses this term without defining it. (It only defines a valuation ring, in this case as a domain $R$ with $\mathcal{Fr}(R)=R\cup R^{-1}$.)
In addition to the above, and probably more importantly, I'd like to ask about motivation. Even if I can understand the step by step technicalities of a proof, I don't see the big picture. Is this problem something that might naturally arise in other mathematical settings? How is the result important? What does it tell us?
 A: Let's start with definitions of valuations.
a) Valuation ring (which sometimes I call it general valuation ring), the definition at the book;
'Algebraic Geometry' written by 'Daniel Bump' page 7: "Let $F$ be a field. A subring $R\subseteq F$ called a valuation ring of $F$ if $0\neq x\in F$ implies that either $x\in R$ or $x^{-1}\in R$. Clearly then $F$ is the field of fractions of $R$."
'Introduction to Commutative Algebra' written by 'M. F. Atiyah' and 'I. G. Macdonald' page 65: "Let $B$ be an integral domain, $K$ its field of fractions. $B$ is a valuation ring of $K$ if, for each $x\neq 0$, either $x\in B$ or $x^{-1}\in B$ (or both)." Which is exactly the definition of the book you were using.
'Undergraduate Commutative Algebra' written by 'Miles Reid' page 116 (the title there is GENERAL VALUATION RINGS): "Let $A$ be an integral domain and $K$=Frac$A$ its field of fractions. Then $A$ is a valuation ring if for every nonzero element $x\in K$, either $x$ or $x^{-1}\in A$."
Again the definition of your book.
'Geometric Aspects of Valuation Theory' from a note taken from lectures of Prof. Teissier: "A valuation ring is an integral domain $V$ such that if $K$ is its field of fraction and we have $x\in K-V$ then $x^{-1}\in V$." Again the definition of your book.
The difference of the first definition and the others is that by the first one which takes $F=$Frac$R$ as a result not part of the definition I am allowed to give the first class of examples in here.
b) Discrete Valuation ring, the definition at the book:
'Algebraic Geometry' written by 'Daniel Bump' page 57: "Let $F$ be a field. By a discrete valuation on $F$, we mean a surjective function
 $\nu : F \twoheadrightarrow \mathbb{Z}\cup\{\infty\}$ with the properties that $\nu(xy)=\nu(x)+\nu(y)$; such that $\nu(x)=\infty$ if and only if $x=0$; and such that $v(x+y)\geq\min \big(\nu(x),\nu(y)\big)$. In this case, it is clear that $A:=\{x\in F\;:\;\nu(x)\geq 0\}$ is a discrete valuation ring of $F$, with maximal ideal $\mathfrak{m}=\{x\in F\;:\;\nu(x)>0\}$. A ring is called a discrete valuation ring if it arises in this way from a discrete valuation on its field of fractions."
c) (I call this valuation ring the generalized version of discrete valuation ring), the definition at:
'Geometric Aspects of Valuation Theory' from a note taken from lectures of Prof. Teissier:  This is same as discrete valuation ring with the only deference that instead of $\mathbb{Z}$ and its usual addition and order we can take every ordered group. Read section "3.1".
You can also find definitions many other places but I wrote those ones I use myself.
Define the map $\nu:k[x,y]\longrightarrow \mathbb{Z}+\mathbb{Z}\alpha$ mapping $\sum c_{n,m}x^ny^m$ to $\min\{n+m\alpha\}$. Then extend it to $k(x,y)$ in this way that $\nu(\dfrac{f(x,y)}{g(x,y)})=\nu\big(f(x,y)\big)-\nu\big(g(x,y)\big)$ ($f(x,y)$ and $g(x,y)$ are prime to each other and $g(x,y)\neq 0$). It is sufficient to show that $\nu|_{k[x,y]}$ has the two required properties then it has them on the whole $k(x,y)$.
Let $f(x,y),f'(x,y)\in k[x,y]$. You can write them as $f(x,y)=c_{n_0,m_0}x^{n_0}y^{m_0}+\sum c_{n,m}x^ny^m$, $f'(x,y)=c'_{n_0',m_0'}x^{n_0'}y^{m_0'}+\sum c'_{n',m'}x^{n'}y^{m'}$ where $\nu(f)=n_0+m_0\alpha$, $\nu(f')=n_0'+m_0'\alpha$ and for other indices $n_0+m_0\alpha<n+m\alpha$, $n_0'+m_0'\alpha<n'+m'\alpha$. Then
$$f(x,y)f'(x,y)=c_{n_0,m_0}c'_{n_0',m_0'}x^{n_0+n_0'}y^{m_0+m_0'}+\sum c_{n,m}c'_{n',m'}x^{n+n'}y^{m+m'}$$
and
$$\left.\begin{array}{l}
n_0+m_0\alpha<n+m\alpha\\
n_0'+m_0'\alpha<n'+m'\alpha
\end{array}\right\}\Longrightarrow n_0+m_0\alpha+n_0'+m_0'\alpha<n+m\alpha+n'+m'\alpha$$
$$\Longrightarrow (n_0+n_0')+(m_0+m_0')\alpha<(n+n')+(m+m')\alpha.$$
Therefore $\nu(ff')=(n_0+n_0')+(m_0+m_0')\alpha=\nu(f)+\nu(f')$.
The second one is obvious because of existence of $\min$ at $\nu$:
$$f+f'=c_{n_0,m_0}x^{n_0}y^{m_0}+\sum c_{n,m}x^ny^m+c'_{n_0',m_0'}x^{n_0'}y^{m_0'}+\sum c'_{n',m'}x^{n'}y^{m'}$$
$$\begin{array}{ccc}
\min\{n_0+m_0\alpha,n_0'+m_0'\alpha\} & \leq & \min(\{n_0+m_0\alpha,n_0'+m_0'\alpha\}\cup\{n+m\alpha\}\cup\{n'+m'\alpha\}\\
& \leq & \min(\{n''+m''\alpha\;:\;c_{n'',m''}\neq 0\text{ in }f+f'\}
\end{array}$$
Now let $A:=\{h\in k(x,y)\;:\;\nu(h)\geq 0\}$. First note that
$$1=1.1\Longrightarrow\nu(1)=\nu(1)+\nu(1)\Longrightarrow \nu(1)=0$$
($\nu(1)\in\mathbb{R}$)
$$\begin{array}{ccc}
\forall h\in k(x,y) & : & hh^{-1}=1\\
& & \nu(h)+\nu(h^{-1})=0\\
& & \nu(h^{-1})=-\nu(h)
\end{array}$$
So if $h\notin k(x,y)$, $\nu(h)<0$ and $\nu(h^{-1})>0$ so $h^{-1}\in A$. Thus we have a valuation ring of $k(x,y)$.
