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From notes an algebra A over K. Has three properties

i) A is an ring under addition and multiplication. ii) A is a vector space over K under addition and scalar multiplication. iii) for all $\alpha \in K$, $x,y \in A$, $(\alpha x)y=x(\alpha y)=\alpha (xy)$

The notes say for an field $K$. $K[X]$ is an algebra over K?

I can't see this from the axiom. Cleary it is a ring. $K[X]$ is a vector space because it closed under scalar multiplication of elements in K.

However, the last iii) I don't see the motivation behind that. Also how is $K[X]$ finite dimensional vector space.

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    The point of iii) is that multiplication from the right by $y$ and multiplication from the left by $x$ should both be $K$-linear.2011-11-26
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    Whenever you define an algebraic structure with multiple operations, there needs to be *some* axiom relating the operations, or it's not a very interesting definition. For example, the definition of a ring is only interesting because of the distributive law -- otherwise the addition and multiplication wouldn't have anything to do with each other. In the present case, axiom (iii) for algebras is important because it requires the ring multiplication and scalar multiplication to be related. Otherwise an algebra would be nothing more than a vector space and a ring with the same additive group.2011-11-26
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    If $c\in K$ is a scalar and $P,Q\in K[X]$ are polynomials, then certainly $(cP)Q=P(cQ)=c(PQ)$, where's the problem? Assuming your definition of ring assumes a unit $1$, property (iii) allows you to identify $c\in K$ with $c.1$, so that your algebra *contains* the field $K$ (and $K$ is commutes with everything: $ac=a(c.1)=c(a.1)=ca$ for all $c\in K$ and $a\in A$).2011-11-26
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    **Hint** $\ $ Eliminating notational abuse it is $\rm\ (c\cdot x) * y = x *(c\cdot y) = c\cdot(x * y)$2011-11-26

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Here is a variation that you might like: A $K$-algebra is a ring homomorphism $f\colon K \to A$ whose image is contained in the centre of $A$. Here $K$ can be any commutative ring. This is more restrictive than your definition, only because rings have identity elements which maps between rings must respect; however, the two examples you've mentioned can be viewed either way.

The example I usually have in mind is that of $n \times n$ matrices with entries in a field $K$. Here $K$ sits inside the algebra as the subring of scalar matrices, and these commute with everything.

Your definition makes no mention of $A$ being finite-dimensional over $K$, and indeed $K[X]$ has an infinite basis $\{1, X, X^2, \ldots\}$.

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    What I worrying about is he says that $\mathbb{H}$ is a $\mathbb{R}$ algebra, but not a $\mathbb{C}$ algebra.2011-11-26
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    @simplicity: *He* who says that?2011-11-26
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    @simplicity: you can easily check that there are many ring homomorphisms $C\to H$ but that none of the *has image contained in the center* of $H$.2011-11-26
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    @simplicity That's a good example. You do have a copy (really, many copies) of $\mathbf C = \mathbf R + \mathbf Ri$ inside of $\mathbf H$, but the elements of that $\mathbf C$ do not commute with everything in the quaternion algebra, as your axiom (iii) requires: $ij = -ji$, for example [Take $\alpha = i$, $x = j$, $y = 1$].2011-11-26
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One simple way to view an $\rm\:R$-algebra $\rm\:A\:$ is simply as a ring $\rm\:A\:$ containing (an image of) $\rm\:R\:$ as a distinguished subring of "coefficients" or "scalars" that multiplicatively commute with all elements of $\rm\:A\:$. This abstracts constructions such as polynomial or matrix rings over the coefficient ring $\rm\:R\:$, or over an image $\rm\:R/I\:$ of $\rm\:R\:$. It is this structure that is being axiomatized above. Since such an extension ring is naturally an $\rm\:R$-module (= $\rm\:R$-vector space when $\rm R$ is a field) one can examine the structure at this level. Doing so requires precisely the axiom (iii) to capture the sought properties of coefficient multiplication. It's an enlightening exercise to prove these two definitions are equivalent.