# Is a field (ring) an algebra over itself?

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!

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What's your definition of algebra? According to mine, a commutative ring is always an algebra over itself. (Unless otherwise indicated to me an $R$-algebra has $R$ commutative). – Robin Chapman Sep 8 '10 at 11:43
Thanks! Updated with links to their definitions. How about a field? – Tim Sep 8 '10 at 12:06
When $R$ is non-commutative, it is not an algebra over itself by the Wikipedia definition, since the second condition of bilinearity there will not be satisfied. – user641 Sep 8 '10 at 12:35

A field K is an algebra over a field (the field being K). A commutative ring R is an algebra over a ring (the ring being R).

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And a not-necessarily-commutative ring $R$ with $1$ is an algebra over its center. – Arturo Magidin Sep 8 '10 at 16:38

[This should be a comment, I guess]

This is the kind of question you can answer yourself by simply checking the conditions in the definition are satisfied. You really should do it, independently of what anyone answers here or elsewhere!

You should know, though, that a field can be an algebra over itself in many different ways, as soon as you have a field $K$ such that there are at least two ring homomorphisms $K\to K$ (there is always one such ring homomorphism, the identity map, so to get something interesting, we need two).

It is easy to give examples of fields which do have such non-identity homomorphisms. For example, every Galois extension of $\mathbb Q$, or the field of rational functions $\mathbb C(x)$ (there is a non-identity homomorphism $\mathbb C(x)\to \mathbb C(x)$ which maps $x$ to $x^2$, and in fact many similar other ones), &c.

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+1: Useful information, but still reminds the poster to try answering the question right from the definition. (Personally, I assume he did, but it would be useful to indicate that in the question, something like, “following the definition, I believe the answer is yes, can someone confirm that?”) – Christopher Creutzig Sep 9 '10 at 12:31
@Christopher, it is rather difficult not to be able to check the definition in this case... – Mariano Suárez-Alvarez Sep 9 '10 at 14:28

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.

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