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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 $K$?

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Is there any compatibility between the ring structure of R and its K-vector space structure? – Mariano Suárez-Alvarez Jul 26 '14 at 2:15
Oh, sure. The sum operation is the same. In the case of my remark about the case $K \subset R$, it is required that ring multiplication and scalar multiplication coincide if an element in $K$ is involved, but in the case $K \subsetneq R$, well, I do not require any compatibility besides addition being the same. – Marco Flores Jul 26 '14 at 4:32
up vote 4 down vote accepted

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.

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It was a while since I last used that algebra as an example! – Mariano Suárez-Alvarez Jul 26 '14 at 6:24

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