# Abstract Nonsense

## R-Algebras

Point of Post: In this post we discuss the notion of $R$-algebras where $R$ is some commutative unital ring, and the associated categories.

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Motivation

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Very often when we run into $R$-modules (where we assume that $R$ is commutative and unital) there is more structure involved, namely the modules are also rings which interact nicely with the $R$-module structure. Namely, we have a (left, right)$R$-module $A$ along with a bilinear map $\cdot:A\times A\to A$ which makes $A$ into a unital ring $A$ such that $r(x\cdot y)=(rx)\cdot y=x\cdot (ry)$. We have already run into algebras before, in the context of endomorphism algebras of vector spaces. More generally, any ring of matrices $\text{Mat}_n(R)$ is given the structure of an $R$-algebra. In fact, it’s easy to see that algebras generalize ring theory since, as we shall see, the category $\mathbb{Z}\text{-}\mathbf{Alg}$ of all $\mathbb{Z}$-algebras is isomorphic to the category $\mathbf{Rng}$ of rings. Other examples of algebras are polynomial algebras and $\mathbb{C}$ is a $2$-dimensional $\mathbb{R}$-algebra. Algebras play an important role in a lot of algebraic subjects, perhaps most notably with their appearance in differential geometry in the form of tensor algebras, and their appearance in commutative algebra.

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$R$-algebras

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Let $R$ be a commutative unital ring, then we call an $R$-module $(A,+)$ (no need for distinction about left or right since $R$ is commutative) together with a $R$-bilinear map $\cdot:A\times A\to A$ such that $(A,+,\cdot)$ is a unital ring, $r(x\cdot y)=(rx)\cdot y=x\cdot(ry)$ and $1_Ra=a$ for all $a\in A$ to be an $R$-algebra.

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There is another common definition of an$R$-algebra. Namely, let $A$ be a unital ring and let $\phi:R\to Z(A)$ be a unital ring homomorphism, we’d like to consider this the same thing as an $R$-algebra.

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Indeed, we can define an $R$-module structure on $A$ by defining $r\cdot a=\phi(r)a$. To see this, we merely note that

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\begin{aligned}&(r+r')\cdot a=\phi(r+r')a=\phi(r)a+\phi(r')a=ra+ra'\\ &r\cdot(a+a')=\phi(r)(a+a')=\phi(r)a+\phi(r)a'\\ & (rr')a=\phi(rr')a=\phi(r)\phi(r')a=\phi(r)(\phi(r')a)=r(r'a)\end{aligned}

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Moreover, this module structure along with the usual ring structure of $A$ makes $A$ into an $R$-algebra. The only thing to check is that $r(aa')=(ra)a'=a(ra')$ but this follows since this can be rewritten as $\phi(r)(aa')=(\phi(r)a)a'=a(\phi(r)a')$ which follows from the associativity of $A$ and the fact that $\phi(R)\subseteq Z(A)$. Moreover, the fact that $A$ is unital follows from the fact that $1_Ra=\phi(1_R)a=1_Aa=a$ for all $a\in A$ from where the conclusion follows.

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Conversely, if $A$ is an $R$-algebra we can define a map $\phi:R\to Z(A)$ by defining $\phi(r)=r\cdot1_A$. To see that this is really a unital $R$-map we note that

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\begin{aligned}&\phi(r+r')=(r+r')1_A=r1_A+r'1_A=\phi(r)+\phi(r')\\ &\phi(rr')=(rr')1_A=(r1_A)(r'1_A)=\phi(r)\phi(r')\\ &\phi(1_R)=1_R1_A=1_A\end{aligned}

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The fact that $\phi(R)\subseteq Z(A)$ follows from the fact that $\phi(r)a=(r1_A)a=r(1_Aa)=r(a1_A)=a(r1_A)=a\phi(r)$ for all $a\in A$ and $r\in R$.

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To consider some examples, we note that for any commutative unital ring $R$ the polynomial rings $R[x_1,\cdots,x_n]$ and the matrix rings $\text{Mat}_n(R)$. In fact, more generally, given any $R$-algebra $A$ we have that $A[x_1,\cdots,x_n]$ and $\text{Mat}_n(A)$ are $R$-algebras with the usual notions of addition and multiplication.

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If $X$ is a topological space then the set $C(X)$ of all continuous maps $X\to\mathbb{R}$ is an $\mathbb{R}$-algebra, similarly the set $C(X;\mathbb{C})$ of all continuous maps $X\to\mathbb{C}$ is a $\mathbb{C}$-algebra.

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Any ring $A$ is a $S$-algebra in the natural way whenever $S\subseteq Z(A)$.

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The Hamiltonian quaternions $\mathbb{H}$ form a $4$-dimensional $\mathbb{R}$-algebra.

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Let $k$ be some field and $V$ a (not necessarily finitely dimensional) $k$-space. Then, the endomorphism algebra $\text{End}_k(V)$ is an algebra over $k$.

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We define a map of $R$-algebras or $R$-algebra homomorphism from the $R$-algebra $A$ to the $R$-algebra $B$ to be a set function $f:A\to B$ which is both a unital ring homomorphism and an $R$-map. If we think about the $R$-algebras $A$ and $B$ as just rings $A$ and $B$ with unital ring homomorphism $\phi:R\to A$ and $\psi:R\to B$. Then, an $R$-algebra map $f:A\to B$ is nothing more than a unital ring map $f:A\to B$ such that $f\circ\phi=\psi$.Indeed, one merely checks that $f\circ\phi=\psi$ implies that

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$f(r\cdot x)=f(\phi(r)x)=f(\phi(r))f(x)=\psi(r)f(x)=r\cdot f(x)$

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so that $f$ is an $R$-map as well.

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As an example of an (iso)morphism of $R$-algebras one can recall the basic fact from linear algebra that if $k$ is a field and $V$ is an $n$-dimensional $k$-space then $\text{End}_k(V)\cong\text{Mat}_n(k)$.

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We then define $R\text{-}\mathbf{Alg}$ to be the category of all $R$-algebras with $R$-algebra homomorphisms. We define $R\text{-}\mathbf{CAlg}$ to be the full subcategory whose objects are $R$-algebras $A$ such that $A$ is commutative. It’s fairly easy to see that there is only one way to define a $\mathbb{Z}$-algebra structure on a given ring $A$, this is because $\mathbb{Z}$ is initial in $\mathbf{Ring}$ and so there is only one unital homomorphism $\mathbb{Z}\to A$. Moreover, it’s easy to verify that this is just the usual $\mathbb{Z}$-module structure on the underlying abelian group of $A$. Thus, we see that $\mathbb{Z}\text{-}\mathbf{Alg}\cong\mathbf{Ring}$ and $\mathbb{Z}\text{-}\mathbf{CAlg}\cong\mathbf{CRing}$.

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[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.

January 10, 2012 -

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