Let $F$ be a field, and $V$ be an $F$-vector space. Make $R=F⊕V$ a ring by putting $xy=0$ for $x,y\in V$. Is it true that the Krull dimension of $R$ is equal to zero?
If this is so, $R$ would be an example of a non-domain of dimension $0$.
Thanks for any help!
On
$\mathbf Z/p^2\mathbf Z$ is the simplest example of a (local) ring of dimension $0$. Its only prime ideal $p\mathbf Z/p^2\mathbf Z$ is the set of nilpotent elements of this ring.
Any product of a finite number of fields has dimension $0$, and is reduced.
On
The ideal $V$ is nilpotent and has $k$ as a quotient, so $V$ is the unique maximal ideal of $k\oplus V$, hence your ring is local with only one prime ideal (if $\mathfrak a\mathfrak b\subseteq \mathfrak p$ and $\mathfrak p$ is prime, then it contains one of the ideals, so for a maximal nilpotent ideal we have $\mathfrak m^n=0\subseteq \mathfrak p$ for every prime, i.e. $\mathfrak m$ is the only prime ideal in your ring). Since any ideal cannot have an element of the form $\lambda + v$ with $\lambda \neq 0$) (since $(\mu+w)(\mu-w)=\mu^2$), every ideal is homogeneous, so every ideal is a subspace of $V$: any subspace of $V$ is a submodule since $V$ acts on itself by $0$. It follows, in particular, that your ring is artinian if and only if it is noetherian if and only if $V$ is finite dimensional.
I think $R$ is the idealization of the $F$-vector space $V$ where the multiplication is given by $$(a,x)(b,y)=(ab,ay+bx),$$ for $a,b\in F$ and $x,y\in V$. The ideal $\{0\}\times V$ is nilpotent since $(\{0\}\times V)^2=0$. Moreover, every element outside of $\{0\}\times V$ is invertible, so $R$ is a local ring with nilpotent maximal ideal. Now it follows easily that $\dim R=0$: if $P$ is a prime ideal of $R$ then $(\{0\}\times V)^2=0\subseteq P$, so $\{0\}\times V\subseteq P$ hence equality.