How to show this ring is Noetherian?

Question

I got the following exercise:
Let $W$ be a finite-dimensional $\Bbb{R}$-vector space. Let $\Bbb{R}_W=\Bbb{R}\times W$. Define addition and multiplication by $(r,w)+(s,v)=(r+s,w+v)$, $(r,w)*(s,v)=(rs,sw+rv)$, for $r,s\in \Bbb{R}$ and $w,v\in W$.
It is easy to show that $\Bbb{R}_W$ is a commutative unitary ring. Now how do I show it is Noetherian? I think I need to find all of its ideals, but I do not know how to do it. Thanks for your help!

Answer 1

I think finding all ideals of $\mathbb{R}_W$ is a good idea. Do not think it too complicated!

Let $I$ be an ideal of $\mathbb{R}_W$. We divide it into two cases.

The first case, $I$ contains an element like $(r, w)$ ($r \neq 0$). Therefore, for every element $(s,x)$ in $\mathbb{R}_W$, we have:

$(r, w) * (\frac{s}{r}, \frac{x}{r} - \frac{s}{r^2}w) = (s, x)$.

Therefore, in the first case, $I = \mathbb{R}_W$ is finitely generated by the identity $(1, 0)$.

The second case, every element in $I$ has the form $(0, w)$. In this case, the finite set $\{(0, w_i)\}$ generates $I$ if the set $\{w_i\}$ forms a base of $W$ (using the condition that $W$ is finite-dimensional).

Answer 2

You can also observe simply that it is an $n+1$ dimensional $\mathbb R$ algebra, where $n$ is the dimension of $W$. That is a hard bound on the length of any chain of ideals, ascending or descending. So it is both Noetherian and Artinian.