Getting a ring from a vector space and a field. Let $k$ be a field and let $V$ be a vector space over $k$. Define the product of two elements in the set $R= k \times V$ by $$(a,v)(a^{'},v^{'})=(aa^{'},av^{'}+a^{'}v)$$ where $a,a^{'} \in k$ and $v,v^{'} \in V$ and the sum by $$(a,v)+(a^{'},v^{'})=(a+a^{'},v+v^{'}).$$I was able to show that $R$ is a ring with multiplicative identity $(1,0)$ and the multiplicative inverse of an element $(a,v)$ is $(\frac{1}{a},\frac{-a'v}{a})$, I hope I am correct? I am difficulty showing the following:


*

*R is a local ring and R has only one prime ideal

*There is a one-to-one correspondence between the set of all ideals $I \subset R$ in $R$ and the set of all vector subspaces in $V$. 
I know that R will be a local ring if it contain a unique maximal ideal, but I can seem to pull it through and for the second part I am guessing the ideals in R will have a certain form that can make create this correspondence. Hints and comments will be highly appreciated.Thanks.
 A: Locality
The thing to note is that $M=\{0\}\times V$ is a nilpotent ideal (it squares to zero) and moreover, $R/M\cong k$. Because it is nilpotent it is contained in all maximal ideals, and because it is maximal, it is the only maximal ideal.
Correspondence
Pick any subspace $W$ of $V$. Consider the additive abelian group $\{0\}\times W\subseteq R$.
Then $(a,v)(0,w)=(0,aw)\in \{0\}\times W$ by definition of multiplication and the fact that $W$ is a subspace. So any subspace of $V$ makes an ideal this way.
Conversely, suppose $I$ is some proper ideal. Obviously it is a vector space under the action $\lambda (a, v):=(\lambda, 0)(a,v)=(\lambda a, \lambda v)$. Moreover, it is a subset of the maximal ideal $M$, so it is actually zero in the left hand coordinates. What you have left is something of the form $\{0\}\times W$ for a subspace $W$ of $V$.
Units
If you have seen $M$ squares to zero, it makes it dramatically easy to see the units: if $a\neq 0$, then $(a,v)(a,-v)=(a^2,0 )$, and by dividing by $a^2$ you get the same formula you came up with for the inverse, $(a,v)^{-1}=\frac{1}{a^2}(a,-v)=(\frac{1}{a},\frac{-v}{a^2})$ .
