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I had previously seen the definition to be the one in Atiyah-Macdonald's Commutative Algebra:

  1. A is a ring and an algebra over a ring is a ring B such that there is a map $\phi:A\rightarrow B$.

which I tried to show is eqivalent to this definition(2) but I could not show that 2 implies 1. In particular, I think that the set of all polynomials over a ring with zero constant term satisfies 2 but not 1. How do I do this?

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It depends on whether you want your rings and algebras to be associative and have a unit. Atiyah-Macdonald wants this. Wikipedia doesn't. Note that if you add $[,]$ is associative and has a unit to (2), it works. – martini Dec 7 '12 at 9:33
Got it. Thanks. – Abhishek Gupta Dec 7 '12 at 9:37
@martini: $\:$ I think that should be an answer. $\;\;$ – Ricky Demer Dec 7 '12 at 9:44
yes indeed, I cannot accept a comment. – Abhishek Gupta Dec 7 '12 at 9:50
@RickyDemer done – martini Dec 7 '12 at 9:50
up vote 1 down vote accepted

The difference between the two definitions of algebra over $A$ is the Atiyah-Macdonald wants all rings to have a unit, all ring homomorphisms to be unitary and all algebras to be associative. The definition from wikipedia does not imply this. If we add this to the definition (2), that is rephrase it as

An $A$-algebra $B$ is an $A$-module $B$ together with an $A$-bilinear associative $[,]\colon B^2\to B$ such that $[,]$ has a unit.

we get the desired $\phi$ by letting $\phi(a) = a1_B$ where the multiplication denotes the $A$-module operation on $B$.

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