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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:

  1. R is a local ring and R has only one prime ideal

  2. 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.

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    Sorry, your proposed multiplicative inverse doesn't work. Check it and see what you need to change (a negative in there somewhere to get 0). – 2017-02-06
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    Yes, That was a typo. Thanks for pointing it out. – 2017-02-06
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    I have corrected it. – 2017-02-06
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    The formula you provided for the inverse of an element clearly only works if $a$ is not zero., OTOH, there is no $a'$ in sight... – 2017-02-06

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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, $(a,v)^{-1}=\frac{1}{a^2}(a,-v)=(\frac{1}{a},\frac{-v}{a^2})$ .

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    Thanks a lot. But how can I develop myself to think really fast this way. – 2017-02-06
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    @Jaynot Well, there are probably a lot of ways to do that, but the one I've found to be most effective is: look for lots of exercises on what you are interested and solve them. Use textbooks or past qualification exams for grad school. Then you get on math.SE and read and write answers, read and write questions, and after a time you pick up a lot of different intuition. I think they key is exposure to different authors' tricks. Reading $100$ different posters' approaches is probably going to benefit you more than reading a single author's textbook, no matter how good it is. – 2017-02-06
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    Thanks. I proved the locality in a different way using your approach. I know that if $R$ is a ring and $M \subset R$ is an ideal such that $R\M$ is a unit. Then $R$ is a local ring and I is a maximal ideal. Here is my proof. Take any element $(a,v) \in R$ such that $a \neq 0$. Then $(a,v) \in R\M$ and observe that elements of this form $(a,v)$ is invertible. Therefore, R is a local ring and the ideal M is a maximal ideal. – 2017-02-07
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    I continued further to say that since every maximal ideal is a prime ideal then the ideal $M$ is a prime ideal of $R$. I am still thinking of the uniqueness part because the question said prove that R contains a unique prime ideal. Once I get it, I will post it here. – 2017-02-07
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    @Jaynot that method you used works to prove it is local. Don't forget that prime ideals contain all nilpotent ideals! – 2017-02-07