Could someone clarify the difference, if any, between an algebra over a field $K$ and a field extension of $K$? Also, can someone provide an example of a finite dimensional algebra over $K$ that has no primitive generator?
About the above question, I know that a finite extension of a field $K$ has a primitive generator, as long as the field is finite or of characteristic zero. Does the same property hold for algebras? (if so, I guess the example must be an infinite field with finite characteristic).
Also, I know that a reduced finite dimensional algebra over a field $K$ of zero characteristic has a primitive generator. What happens when instead $K$ is finite? Could someone provide an example of such an algebra which has no primitive generator?
I can't answer all of your questions, but I can tell you this much: A $K$ algebra is a ring $R$ with a ring homomorphism $\varphi : K \to R$. Thus you can multiply each element of $R$ by any element of $K$ via $k\cdot r = \varphi(k) r$.
Generally, one might not actually assume that an algebra is a ring, in which case a $K$ algebra is simply a $K$-vector space with a bilinear multiplication map satisfying no additional assumptions.
By contrast, a field extension $L$ of $K$ is a field which contains $K$ as a subfield. Thus, it is a very special case of an algebra.
Yes, the notation $(x, y)^2$ means the product ideal $(x, y) \cdot (x, y)$. It's generated by the elements $x^2, xy, yx, y^2$.