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Homelinearization
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linearization
Linearization is the process of reducing a homogeneous polynomial into a multilinear map over a commutative ring. There are in general two ways of doing this:

Method 1. Given any homogeneous polynomial $f$ of degree $n$ in $m$ indeterminates over a commutative scalar ring $R$ (scalar simply means that the elements of $R$ commute with the indeterminates).
 Step 1
If all indeterminates are linear in $f$, then we are done.
 Step 2
Otherwise, pick an indeterminate $x$ such that $x$ is not linear in $f$. Without loss of generality, write $f=f(x,X)$, where $X$ is the set of indeterminates in $f$ excluding $x$. Define $g(x_{1},x_{2},X):=f(x_{1}+x_{2},X)f(x_{1},X)f(x_{2},X)$. Then $g$ is a homogeneous polynomial of degree $n$ in $m+1$ indeterminates. However, the highest degree of $x_{1},x_{2}$ is $n1$, one less that of $x$.
 Step 3
Repeat the process, starting with Step 1, for the homogeneous polynomial $g$. Continue until the set $X$ of indeterminates is enlarged to one $X^{{{}^{{\prime}}}}$ such that each $x\in X^{{{}^{{\prime}}}}$ is linear.
 Step 1

Method 2. This method applies only to homogeneous polynomials that are also homogeneous in each indeterminate, when the other indeterminates are held constant, i.e., $f(tx,X)=t^{n}f(x,X)$ for some $n\in\mathbb{N}$ and any $t\in R$. Note that if all of the indeterminates in $f$ commute with each other, then $f$ is essentially a monomial. So this method is particularly useful when indeterminates are noncommuting. If this is the case, then we use the following algorithm:
 Step 1
If $x$ is not linear in $f$ and that $f(tx,X)=t^{n}f(x,X)$, replace $x$ with a formal linear combination of $n$ indeterminates over $R$:
$r_{1}x_{1}+\cdots+r_{n}x_{n}\mbox{, where }r_{i}\in R.$  Step 2
Define a polynomial $g\in R\langle x_{1},\ldots,x_{n}\rangle$, the noncommuting free algebra over $R$ (generated by the noncommuting indeterminates $x_{i}$) by:
$g(x_{1},\ldots,x_{n}):=f(r_{1}x_{1}+\cdots+r_{n}x_{n}).$  Step 3
 Step 4
Take the next nonlinear indeterminate and start over (with Step 1). Repeat the process until $f$ is completely linearized.
 Step 1
Remarks.
1. The method of linearization is used often in the studies of Lie algebras, Jordan algebras, PIalgebras and quadratic forms.
2. If the characteristic of scalar ring $R$ is 0 and $f$ is a monomial in one indeterminate, we can recover $f$ back from its linearization by setting all of its indeterminates to a single indeterminate $x$ and dividing the resulting polynomial by $n!$:
$f(x)=\frac{1}{n!}\operatorname{linearization}(f)(x,\ldots,x).$ Please see the first example below.
3. If $f$ is a homogeneous polynomial of degree $n$, then the linearized $f$ is a multilinear map in $n$ indeterminates.
Examples.

$f(x)=x^{2}$. Then $f(x_{1}+x_{2})f(x_{1})f(x_{2})=x_{1}x_{2}+x_{2}x_{1}$ is a linearization of $x^{2}$. In general, if $f(x)=x^{n}$, then the linearization of $f$ is
$\sum_{{\sigma\in S_{n}}}x_{{\sigma(1)}}\cdots x_{{\sigma(n)}}=\sum_{{\sigma\in S% _{n}}}\prod_{{i=1}}^{{n}}x_{{\sigma(i)}},$ where $S_{n}$ is the symmetric group on $\{1,\ldots,n\}$. If in addition all the indeterminates commute with each other and $n!\neq 0$ in the ground ring, then the linearization becomes
$n!x_{1}\cdots x_{n}=\prod_{{i=1}}^{{n}}ix_{i}.$ 
$f(x,y)=x^{3}y^{2}+xyxyx$. Since $f(tx,y)=t^{3}f(x,y)$ and $f(x,ty)=t^{2}f(x,y)$, $f$ is homogeneous over $x$ and $y$ separately, and thus we can linearize $f$. First, collect all the monomials having coefficient $abc$ in $(ax_{1}+bx_{2}+cx_{3},y)$, we get
$g(x_{1},x_{2},x_{3},y):=\sum x_{i}x_{j}x_{k}y^{2}+x_{i}yx_{j}yx_{k},$ where $i,j,k\in{1,2,3}$ and $(ij)(jk)(ki)\neq 0$. Repeat this for $y$ and we have
$h(x_{1},x_{2},x_{3},y_{1},y_{2}):=\sum x_{i}x_{j}x_{k}(y_{1}y_{2}+y_{2}y_{1})+% (x_{i}y_{1}x_{j}y_{2}x_{k}+x_{i}y_{2}x_{j}y_{1}x_{k}).$
Mathematics Subject Classification
15A63 no label found15A69 no label found16R99 no label found17A99 no label found Forums
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