Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [algebras]

For questions about algebras, their properties, and their structures. Use [tag:algebra-precalculus] or [tag:abstract-algebra] if your question is about algebra, not algebras.

An algebra over a field is a vector space equipped with a bilinear product. This product is not necessarily associative or unital, but if it is then the algebra is also a ring with unity. This can also be generalized by assuming that the scalars come from a commutative ring, rather than a field.

As with other algebraic objects, it is possible to define algebra homomorphisms, subalgebras, ideals, etc.

Examples

  • Group algebras, the algebra of polynomials $K[x]$ over a field $K$, and the quaternions are all associate algebras.

  • Every ring is an associative algebra over it's center.

  • The octonions are a non-associative algebra, and Lie algebras may not be associative.

Finite-dimensional algebras can be classified up to isomorphism by selecting a basis of $n$ and describing the multiplication of any two basis elements, which requires $n^3$ structure coefficients.

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What is the difference between a Ring and an Algebra?

In mathematics, I want to know what is indeed the difference between a ring and an algebra?
user70795
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Proof the quaternions are 4-dimensional?

The quaternions can be defined as $$\mathbb{R}\langle X,Y\rangle/(X^2+1,Y^2+1,XY+YX)$$ From these relations, it is relatively easy to prove that $1,X,Y,XY$ span the quaternions over $\mathbb{R}$. But I cannot find any way to prove that this is a…
Zoe Allen
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Are all finite-dimensional algebras of a fixed dimension over a field isomorphic to one another?

Suppose I have a finite-dimensional algebra $V$ of dimension $n$ over a field $\mathbb{F}$. Then $V$ is an $n$-dimensional vector space and comes equipped with a bilinear product $\phi : V \times V \to V$. Suppose now that I have another…
Perturbative
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Does the product rule imply the chain rule?

Let $\mathbb{F}$ be a field, and consider $\mathbb{F}^\mathbb{F}$ as an algebra over $\mathbb{F}$ with the standard function multiplication. Let $D$ be a derivation on a subalgebra of $\mathbb{F}^\mathbb{F}$ closed under function composition that…
mathlander
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What is an $R$-algebra?

In the following, assume that rings are rings with unity. Here is the definition of $R$-algebra from Wikipedia: Let $R$ be a commutative ring and $(M,+,\cdot)$ an $R$-module. Assume $\ast$ is a binary operation on $M$, such that: $x\ast (y+z)=…
Rubertos
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What is an example of a smooth function in $C^\infty(\mathbb{R}^2)$ which is not contained in $C^\infty(\mathbb{R})\otimes C^\infty(\mathbb{R})$

When looking at the tensor product of the ring of smooth functions on $\mathbb{R}^n$, there is only an injection $$ C^\infty(\mathbb{R}^n)\otimes_\mathbb{R}C^\infty(\mathbb{R}^m) \to C^\infty(\mathbb{R}^{n+m}) $$ This motivates the construction of…
54321user
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Can non-isomorphic abelian groups have isomorphic endomorphism rings?

I am aware that distinct Banach spaces $X$, $Y$, give rise to distinct operator algebras $B(X)$, $B(Y)$, but the proof seems to rely heavily on the use of projections and the Hahn-Banach theorem. So if there is a generalization to rings and groups,…
Chrystomath
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Representation theory approach VS Module theory approach?

Given an associative algebra $A$, there is a correspondence between representations of $A$ and left $A-$ modules. Thus, one can study the representation theory of an associative algebra via its left modules and vice versa. My question is that, what…
Math137
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Twisted group algebra coincides with the group algebra?

Let $\mathbb{F}$ be an arbitrary field and $G$ a finite abelian group. Then we can construct the group algebra $\mathbb{F}G$, where, under some conditions on the base field, it will be isomorphic to $\mathbb{F}^{|G|}$ (sum of $|G|$ copies of…
FYY
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Ring homomorphism of tensor product of algebras

Let $B, C$ be two $A$-algebras, $f:A \to B, g: A\to C$ the corresponding ring homomorphisms. From this we can construct an $A$-algebra $B \otimes _A C$ and the mapping $ a \mapsto f(a) \otimes g(a)$ is the corresponding ring homomorphism $A \to B…
Jeong
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What is the difference between a calculus and an algebra?

You can have a lambda calculus, the calculus of the real numbers or a logical calculus but on the other hand you could also have an algebra of sets, a Lie algebra, or a linear algebra. Is there any convention which dictates whether something is an…
Christian Chapman
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Does this property of algebra morphisms (related to idempotents) have a name?

Let $F$ be a field. I am in the category of finite-dimensional $F$-algebras. Let $f:A \rightarrow B$ a homomorphism of two of those. The property of $f$ which came up as useful in something I consider is: $$(*) \text{ For every idempotent } e \in A,…
Torsten Schoeneberg
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Field which is a finitely generated algebra over a field k, is finite over k

Let $k$ be a field and $L$ a finitely generated $k$-algebra, which is also a field. Then $L$ is finite over $K$ Finitely generated means $L = k[x_1, \dots, x_n]$ ($x_i \in L$), and finite over $k$ means $L$ is a finitely generated $k$-module. I am…
chilliBeanDream
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Show that $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ for $n>0$

This is an exercise from Hatcher's Algebraic Topology (exercise 2.B.8). Here is the problem statement: Show that, for $n>0$, $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ by showing that if it were, then for nonzero $a \in \Bbb R^{2n+1}$…
G Pace
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Where does Gelfand Theory fail for non-commutative algebras.

I'm trying to get my head around Gelfand theory, and I can't seem to find the subtleties between commutative and non-commutative algebras. Why is there not a one-to-one correspondence between maximal ideals of a non-commutative algebra and the…
Jacob Denson
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