883: Algebra over Field

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definition of algebra over field

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Topics

About: algebra over field

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of %field name% vectors space.

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Target Context

• The reader will have a definition of algebra over field.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$\ast A$: $\in \{\text{ the vectors spaces over }F\}$, with any multiplication, $\bullet :A\times A\to A$, specified below
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Conditions:
$\mathrm{\forall }{r}_1,r_2,r_1^{\prime },r_2^{\prime }\in F,\mathrm{\forall }{v}_1,v_2,v_1^{\prime },v_2^{\prime }\in A((r_1{v}_1+r_2{v}_2)\bullet (r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })+(r_1{r}_2^{\prime })(v_1\bullet v_2^{\prime })+(r_2{r}_1^{\prime })(v_2\bullet v_1^{\prime })+(r_2{r}_2^{\prime })(v_2\bullet v_2^{\prime }))$
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2: Note

In other words, $\bullet$ is bilinear.

Inevitably, for each $v\in A$, $v\bullet 0=0\bullet v=0$, because $v\bullet 0=(v+0)\bullet (0+0)=(1v+0v)\bullet (0v+0v)=(10)(v\bullet v)+(10)(v\bullet v)+(00)(v\bullet v)+(00)(v\bullet v)=0(v\bullet v)+0(v\bullet v)+0(v\bullet v)+0(v\bullet v)=0+0+0+0=0$ and $0\bullet v=(0+0)\bullet (v+0)=(0v+0v)\bullet (1v+0v)=(01)(v\bullet v)+(00)(v\bullet v)+(01)(v\bullet v)+(00)(v\bullet v)=0(v\bullet v)+0(v\bullet v)+0(v\bullet v)+0(v\bullet v)=0+0+0+0=0$.

So, the requirement of this definition equals this set of conditions: 1) $(v_1+v_2)\bullet v_1^{\prime }=v_1\bullet v_1^{\prime }+v_2\bullet v_1^{\prime }$; 2) $v_1\bullet (v_1^{\prime }+v_2^{\prime })=v_1\bullet v_1^{\prime }+v_1\bullet v_2^{\prime }$; 3) $r_1{v}_1\bullet r_1^{\prime }{v}_1^{\prime }=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })$: supposing the condition of this proposition, 1) $(v_1+v_2)\bullet v_1^{\prime }=(1v_1+1v_2)\bullet (1v_1^{\prime }+0)=1v_1\bullet v_1^{\prime }+1v_1\bullet 0+1v_2\bullet v_1^{\prime }+1v_2\bullet 0=1v_1\bullet v_1^{\prime }+0+1v_2\bullet v_1^{\prime }+0=v_1\bullet v_1^{\prime }+v_2\bullet v_1^{\prime }$; 2) $v_1\bullet (v_1^{\prime }+v_2^{\prime })=(1v_1+0)\bullet (1v_1^{\prime }+1v_2^{\prime })=1v_1\bullet v_1^{\prime }+1v_1\bullet v_2^{\prime }+0\bullet 1v_1^{\prime }+0\bullet 1v_2^{\prime }=1v_1\bullet v_1^{\prime }+1v_1\bullet v_2^{\prime }+0+0=v_1\bullet v_1^{\prime }+v_1\bullet v_2^{\prime }$; 3) $r_1{v}_1\bullet r_1^{\prime }{v}_1^{\prime }=(r_1{v}_1+0)\bullet (r_1^{\prime }{v}_1^{\prime }+0)=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })+r_1{v}_1\bullet 0+0\bullet r_1^{\prime }{v}_1^{\prime }+0\bullet 0=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })+0+0+0=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })$; supposing the conditions, 1), 2), and 3), $(r_1{v}_1+r_2{v}_2)\bullet (r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })=r_1{v}_1\bullet (r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })+r_2{v}_2\bullet (r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })=r_1{v}_1\bullet r_1^{\prime }{v}_1^{\prime }+r_1{v}_1\bullet r_2^{\prime }{v}_2^{\prime }+r_2{v}_2\bullet r_1^{\prime }{v}_1^{\prime }+r_2{v}_2\bullet r_2^{\prime }{v}_2^{\prime }=(r_1{r}_1^{\prime })(v_1\bullet v_1^{\prime })+(r_1{r}_2^{\prime })(v_1\bullet v_2^{\prime })+(r_2{r}_1^{\prime })(v_2\bullet v_1^{\prime })+(r_2{r}_2^{\prime })(v_2\bullet v_2^{\prime })$.

This definition does not require the associativity of the multiplication, which another definition requires. The reason is that if the associativity was required, "Lie algebra" would be a false name.

An algebra over a field may not be any ring, because the associativity may not hold and any multiplicative identity, $1$, may not exist in it.

Any algebra satisfying the associativity is called "associative algebra", which some people call "algebra".

Any algebra with $1$ is called "unitary algebra".

Any unitary associative algebra is a ring.

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References

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