Let $K$ be a field and $R$ a ring with finite dimensional vector space structure over $K$. Is $R$ necessarily a Noetherian ring?
If $K \subset R$, then any ideal in $R$ is also a subspace and, since a finite basis for it exists, it is a finitely generated ideal, and therefore $R$ is Noetherian.
But what if $R$ does not contain $K$?
Let $k$ be a field and let $K$ be any extension of $k$ of infinite degree —for example, we could take $K=k(t)$, the field of rational functions with coefficients in $k$ in the variable $t$. Then $K$ is an infinite dimensional $k$-vector space, so there is a $k$-basis $B=\{x_i:i\in I\}$ for some infinite set $I$. Suppose $i_0\in I$ and let $J=I\setminus\{i_0\}$. There is a unique unital $k$-algebra structure on $K$ such that $x_{i_0}$ is the unit and $x_ix_j=0$ for all $i$, $j\in J$. As $I$ is infinite, this algebra structure on $K$ is not noetherian.
Now clearly $K$ is a finite dimensional $K$-vector space and, with respect to the same addition, it is a non-noetherian algebra as described.