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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!

J126
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Tim
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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 08 '10 at 11:43
  • Thanks! Updated with links to their definitions. How about a field? – Tim Sep 08 '10 at 12:06
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    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. –  Sep 08 '10 at 12:35

3 Answers3

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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).

alext87
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    And a not-necessarily-commutative ring $R$ with $1$ is an algebra over its center. – Arturo Magidin Sep 08 '10 at 16:38
  • @ArturoMagidin Is the fact that a non-commutative ring $R$ is not an $R$-algebra the consequence of the compatibility of the scalar product as an $R$-module? I.e., is this correct: if $R$ is an $R$-algebra (where $\cdot: R^2 \to R$ denotes scalar product defined as ring multiplication, $: R^2 \to R$ is algebra multiplication also defined as ring multiplication), then for all $x, y \in R$, we have $xy = x y = (1 \cdot x) * (y \cdot 1) = (1y) \cdot (x * 1) = y \cdot x = yx$ – Anakhand Feb 18 '22 at 09:31
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[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.

Mariano Suárez-Álvarez
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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 09 '10 at 12:31
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    @Christopher, it is rather difficult not to be able to check the definition in this case... – Mariano Suárez-Álvarez Sep 09 '10 at 14:28
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    @MarianoSuárez-Álvarez Sometimes it is not about that. Especially for those students who are very new to a concept, sometimes they just either lack of confidence or not very sure something so trivial is true. This is sometimes due to the habit of lectures/text book that always make claims that should be verified (or at least tell the reader to verify) look easy without mentioning "Reader should verify directly that ...". This is so common in so-called graduate text books that assume too much instead of making definitions/propositions/proofs accessible. – 10understanding Mar 12 '20 at 13:15
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    @MarianoSuárez-Álvarez (continued) That habit kinda makes these students try to be so quick in concluding some facts (as that way is what they usually read/listen too), which can result in this unsure feeling to any trivial-looking statement, since (once again emphasized) lecturers/textbooks should at least persuade the students to verify instead of concluding something out of nowhere to look fast and smart (or for brevity), even though it is true. Some texts (yes, only texts) in textbooks does this, but most textbooks I have read/have been recommended to read don't. – 10understanding Mar 12 '20 at 13:19
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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.

Bill Dubuque
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