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Homesubalgebra of an algebraic system

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# subalgebra of an algebraic system

Let $(A,O)$ be an algebraic system ($A\neq\varnothing$ is the underlying set and $O$ is the set of operators on $A$).

Subalgebras of an Algebra

Let $B$ be a non-empty subset of $A$. $B$ is *closed* under operators of $A$ if for each $n$-ary operator $\omega_{A}$ on $A$, and any $b_{1},\ldots,b_{n}\in B$, we have $\omega_{A}(b_{1},\ldots,b_{n})\in B$.

Suppose $B$ is closed under operators of $A$. For each $n$-ary operator $\omega_{A}$ on $A$, we define $\omega_{B}:B^{n}\to B$ by $\omega_{B}(b_{1},\ldots,b_{n}):=\omega_{A}(b_{1},\ldots,b_{n})$. Each of these operators is well-defined and is called a *restriction* (of the corresponding $\omega_{A}$). Furthermore, $(B,O)$ is a well-defined algebraic system, and is called the *subalgebra* of $(A,O)$. When $(B,O)$ is a subalgebra of $(A,O)$, we also say that $(A,O)$ is an *extension* of $(B,O)$.

$(A,O)$ is clearly a subalgebra of itself. Any other subalgebra of $(A,O)$ is called a *proper subalgebra*.

Remark. If $(A,O)$ contains constants, then any subalgebra of $(A,O)$ must contain the exact same constants. For example, the ring $\mathbb{Z}$ of integers is an algebraic system with no proper subalgebras. Indeed, if $R$ is a subring of $\mathbb{Z}$, $1\in R$, so $R=\mathbb{Z}$.

Since we are operating under the same operator set, we can, for convenience, drop $O$ and simply call $A$ an algebra, $B$ a subalgebra of $A$, etc… If $B_{1},B_{2}$ are subalgebras of $A$, then $B_{1}\cap B_{2}$ is also a subalgebra. In fact, given any set of subalgebras $B_{i}$ of $A$, their intersection $\bigcap B_{i}$ is also a subalgebra.

Generating Set of an Algebra

Let $C$ be any subset of an algebra $A$. Consider the collection $[C]$ of all subalgebras of $A$ containing $C$. This collection is non-empty because $A\in[C]$. The intersection of all these subalgebras is again a subalgebra containing the set $C$. Denote this subalgebra by $\langle C\rangle$. $\langle C\rangle$ is called the subalgebra *spanned* by $C$, and $C$ is called the *spanning set* of $\langle C\rangle$. Conversely, any subalgebra $B$ of $A$ has a spanning set, namely itself: $B=\langle B\rangle$.

Given a subalgebra $B$ of $A$, a minimal spanning set $X$ of $B$ is called a *generating set* of $B$. By minimal we mean that the set obtained by deleting any element from $X$ no longer spans $B$. When $B$ has a generating set $X$, we also say that $X$ *generates* $B$. If $B$ can be generated by a finite set, we say that $B$ is *finitely generated*. If $B$ can be generated by a single element, we say that $B$ is *cyclic*.

Remark. $\langle\varnothing\rangle=$ the subalgebra generated by the constants of $A$. If no such constants exist, $\langle\varnothing\rangle:=\varnothing$.

From the discussion above, the set of subalgebras of an algebraic system forms a complete lattice. Given subalgebras $A_{i}$, $\bigvee A_{i}$ is the intersection of all $A_{i}$, and $\bigvee A_{i}$ is the subalgebra $\langle\bigcup A_{i}\rangle$. The lattice of all subalgebras of $A$ is called the *subalgebra latttice* of $A$, and is denoted by $\operatorname{Sub}(A)$.

## Mathematics Subject Classification

08A30*no label found*08A05

*no label found*08A62

*no label found*

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