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In mathematics, I want to know what is indeed the difference between a ring and an algebra?

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    $\begingroup$ Multiplication in an algebra is much looser than that in a ring. $\endgroup$ Commented May 3, 2013 at 11:16
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    $\begingroup$ An algebra is a combination of vector space (or module) and ring where all operations are compatible. $\endgroup$
    – lhf
    Commented May 3, 2013 at 11:27
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    $\begingroup$ A unital associative ring is a $\mathbb Z$-algebra. $\endgroup$ Commented May 3, 2013 at 13:39

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A ring $R$ has operations $+$ and $\times$ satisfying certain axioms which I won't repeat here. An (associative) algebra $A$ similarly has operations $+$ and $\times$ satisfying the same axioms (it doesn't need a multiplicative identity, but this axiom isn't always assumed in rings either), plus an additional operation $\cdot\;\colon R\times A\to A$, where $R$ is some ring (often a field) that satisfies some axioms making it compatible with the multiplication and addition in $A$. You should think of this as an analogue of scalar multiplication in vector spaces.

Note also that there are non-associative algebras, so the axioms on multiplication can be weakened from those in rings.

As a vague summary, the algebraic structure of a ring is entirely internal, but in an algebra there is also structure coming from interaction with an external ring of scalars.

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    $\begingroup$ Good answer: I would just add that that the ring of scalars is internal in many cases. An algebra with 1 over a field, for example, contains a copy of the field, and the scalar action matches the ring action. More generally, given a commutative ring $R$ and ring $A$, $A$ becomes an $R$ algebra via any homomorphism $R\to Cen(A)$. The action of the image of this homomorphism is a "close to internal" description of the action of $R$ on $A$. $\endgroup$
    – rschwieb
    Commented May 3, 2013 at 13:17
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    $\begingroup$ @rschwieb This is a good observation, and the inclusion of $R$ into $A$ via pairing the identities sounds like it would be canonical in some appropriate sense; it certainly matches the $R$-action with the action of $R\cdot1\subset A$ on $A$. $\endgroup$
    – mdp
    Commented May 3, 2013 at 15:56
  • $\begingroup$ @MattPressland you have not assumed the ring R is commutative.So it it not required?second question is many people give definition of algebra using bi-linear map(wiki)...how these two are equivalent? $\endgroup$
    – Ripan Saha
    Commented Dec 14, 2014 at 20:29
  • $\begingroup$ @RipanSaha Rings in general need not be commutative, but algebras should be over commutative rings. The bilinear map referred to on Wikipedia is the operation $\times$ - the fact that it is bilinear (over $R$) is one of the compatibility axioms between the multiplication in $A$ and the multiplication with elements of $R$. $\endgroup$
    – mdp
    Commented Dec 19, 2014 at 22:51
  • $\begingroup$ so, can we say informally that an algebra is a ring with a "scalar multiplication" ? $\endgroup$
    – SiXUlm
    Commented Feb 22, 2016 at 17:32
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One thing that complicates answering this question is that rings are almost always assumed to be associative, but algebras are frequently not assumed to be associative. (In other words, my impression is that it's more common to allow 'algebra' to name something nonassociative than it is to use 'ring' to mean something nonassociative.)

Nonassociative algebras are not rare: Lie algebras and Jordan algebras are common nonassociative algebras.

Associative algebras are not rare either: Every single ring $R$ is an associative algebra over its center!

Both structures might or might not be defined to have an identity, so we'll just overlook that feature.

Here's my take, (even though I think Matt Pressland's answer is pretty good already.) $R$ is a commutative ring.

An associative $R$-algebra $A$ is certainly a ring, and a nonassociative algebra may still be counted as a nonassociative ring.

The extra ingredient is an $R$ module structure on $A$ which plays well with the multiplication in $A$. (This was well described before by Matt P: indeed, they are like "scalars".)

In a nutshell, that module action and compatilibity is described by a ring homomorphism from $R$ into the center of $End(A)$, the ring of additive endomorphisms of $A$.

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For a commutative ring $k$, a $k$-algebra is a ring $A$ together with an extra datum: a homomorphism from $k$ into the center of $A$.

The definition allows non-associative algebras if you allow non-associative rings. The most well-understood case occurs when $k$ is a field, and the right way to think about the general case is as an algebraic family of algebras over fields---one for each prime ideal of $k$.

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In the contexts I'm used to, rings are defined to have a constant $1$ and the corresponding axioms making it a multiplicative unit. Algebras, however, do not.

And while you can talk about (for some fixed ring $R$) "rings over $R$" just as you can "algebras over $R$", the latter phrase is far more common than the former.

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    $\begingroup$ I see the same definition in mathworld.wolfram.com/Ring.html. I do not understand why you are downvoted. Seems like holy wars - some bigots dislike to hear your definition :) $\endgroup$
    – Val
    Commented Jul 12, 2014 at 19:03

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