TITLE: Is a field (ring) an algebra over itself?
QUESTION [8 upvotes]: I was wondering if a field is an algebra over itself (http://en.wikipedia.org/wiki/Algebra_over_a_field)?
Also is a ring an algebra over itself (http://en.wikipedia.org/wiki/Algebra_(ring_theory)? If not, does the ring require to be commutative?
Thanks and regards!

REPLY [5 votes]: The answer is clear if you know the universal view of $\rm\:R\:$-algebras, namely they are simply those rings $\rm\:A\:$ into which one can always evaluate $\rm R[x]$ for every $\rm\:a \in A\:$. More directly,  an $\rm\:R$-algebra is just a ring $\rm A\:$ containing a central subring $\rm R'$ that's a ring image of $\:\rm R\:,\:$ i.e. $\rm\: R'\:$ is either an embedding of $\rm\:R\:$ or $\rm\:R/I\:$ for some ideal $\rm\;I\in R\:.\;$ Being central is precisely the condition needed for elts in $\rm\:R'\:$ to serve as "coefficients" in the sense that this makes polynomial rings $\rm\: R[x]\:$ be a universal $\rm\:R\:$-algebras. Namely, the fact that the coefficients commute with all elts of $\rm\:A\:$ is precisely what is required to make the evaluation map be a ring homomorphism $\rm\: R[x]\to A\:,\:$
viz $\rm\;\; r\; x = x\: r\;$ in $\rm\:R[x]\;\Rightarrow\;  r\: a = a\:r\;\;$ by evaluation $\rm\: x\to a\in A,\:$ i.e. by definition polynomial multiplication assumes that the coefficients commute with the indeterminates, so this property must remain true at values of indeterminates if evaluation is to be a ring homomorphism.
A simple but powerful application: $ $  factorizations of polynomials persist as operator factorizations ("operator algebra"). when we evaluate them into $R$-algebra of linear operators, e.g. matrix algebras, and linear differential and difference operators (recurrences), etc.