Ring as a finite dimensional Vector space over a field K

Question

If R is a commutative ring containing a field K as subring and assume that R is finite dimensional vector space over K then show that every prime ideal of R will be a maximal ideal.

My approach was to show for any prime ideal I, R/I will be a finite integral domain and hence a field which will result in I being maximal. But, here it is not given that K is finite field. Since I have just started studying this course and don't know much about Artinian or Noetherian ring so can anyone please solve this problem for infinite field using very basic ideas of ring & vector space theory and not going into domain of Artinian or Noetherian ring.

score 5 · Accepted Answer · answered May 16 '17 at 16:09

5

The quotient $R/I$ will be an integral domain that is finite dimensional over $K$. Any integral domain $S$, which is a finite-dimensional algebra over a field $K$ is a field. To prove this, let $a$ be a nonzero element of $S$. The map $\mu:x\mapsto ax$ is a linear transformation on the vector space $S$ over $K$. By integral domainness, $\mu$ is injective, so by finite-dimensionality, it is also surjective: there is $x\in S$ with $ax=1$. So $S$ is a field.

So $R/I$ is a field: $I$ is maximal.

answered May 16 '17 at 16:09

Angina Seng

158,341

Thanks. BUT How finite dimensionality implies surjectiveness given the field is infinite – Shak May 16 '17 at 16:58
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@ShashJ This is just linear algebra: the rank-nullity formula. – Angina Seng May 16 '17 at 16:59

Ring as a finite dimensional Vector space over a field K

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