## Group Theory and Linear Algebra

Last updated: 27 September 2014

## Lecture 36: Revision: $ℂ\left[t\right]\text{-modules}$

(1) Let $A= ( 123 456 987 ) .$ $A$ acts on $ℂ3={(a1a2a3) | a1,a2,a3∈ℂ}.$ ${ℂ}^{3}$ has basis $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ where ${e}_{1}=\left(1\\ 0\\ 0\\ \right),$ ${e}_{2}=\left(0\\ 1\\ 0\\ \right),$ ${e}_{3}=\left(0\\ 0\\ 1\\ \right)\text{.}$ Define an action of $ℂ\left[t\right]$ on ${ℂ}^{3}$ by, for example, $(3+4t+5t2) (312)= (3+4A+5A2) (312).$ This is the $ℂ\left[t\right]\text{-module}$ defined by $A\text{.}$

(2) Let $ℂ[t]⊕3= { (g1g2g3) | g1,g2,g3∈ℂ[t] }$ so that, for example, $( 3t+7 t3-2t+1 5t2+t6 ) ∈ℂ[t]⊕3 and(312) ∈ℂ[t]⊕3.$ $ℂ\left[t\right]$ acts on $ℂ{\left[t\right]}^{\oplus 3}$ by, for example, $(3+4t+5t2) (312)= ( 9+12t+15t2 3+4t+5t2 6+8t+10t2 ) .$ $ℂ{\left[t\right]}^{\oplus 3}$ has basis: ${ (100), (010), (001), (t00), (0t0), (00t), (t200), (0t20), (00t2), … } .$ $ℂ{\left[t\right]}^{\oplus 3}$ is infinite dimensional. $ℂ{\left[t\right]}^{\oplus 3}$ is the free $ℂ\left[t\right]\text{-module}$ generated by $\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}\text{.}$ A $ℂ\left[t\right]\text{-module}$ is a vector space $V$ over $ℂ$ with a function $ℂ[t]×V ⟶ V (g,v) ⟼ gv$ such that

 (a) If ${g}_{1},{g}_{2}\in ℂ\left[t\right]$ and $v\in V$ then $\left({g}_{1}{g}_{2}\right)v={g}_{1}\left({g}_{2}v\right),$ (b) If $v\in V$ then $1·v=v,$ (c) If ${c}_{1},{v}_{2}\in ℂ,$ ${g}_{1},{g}_{2}\in ℂ\left[t\right]$ and $v\in V$ then $(c1g1+c2g2)v= c1g1v+c2g2v,$ (d) If ${c}_{1},{c}_{2}\in ℂ,$ $g\in ℂ\left[t\right]$ and ${v}_{1},{v}_{2}\in V$ then $g(c1v1+c2v2)= c1gv1+c2gv2.$

A $ℂ\left[t\right]\text{-module}$ homomorphism from $V$ to $W$ is a linear transformation $f:V\to W$ such that if $g\in ℂ\left[t\right]$ and $v\in V$ then $f(gv)=gf(v).$

The kernel of $f$ is $ker f={v∈V | f(v)=0}$ and the image of $f$ is $im f={f(v) | v∈V}.$

(3) The map $Φ: ℂ[t]⊕3 ⟶ ℂ3 (g1g2g3) ⟼ g1(A)(100)+ g2(A)(010)+ g3(A)(001)$ is a $ℂ\left[t\right]\text{-module}$ homomorphism.

$\text{ker} \mathrm{\Phi }=\left(t-A\right)ℂ{\left[t\right]}^{\oplus 3}\text{.}$

More specifically, $t-A = ( t-1-2-3 -4t-5-6 -9-8t-7 ) and ker Φ = { ( t-1-2-3 -4t-5-6 -9-8t-7 ) ( g1(t) g2(t) g3(t) ) | g1(t), g2(t), g3(t) } .$

Consequence (1st homomorphism theorem): $ℂ3≃ ℂ[t]⊕3 (t-A)ℂ[t]⊕3$ as $ℂ\left[t\right]\text{-modules.}$

POINT: $A$ acting in ${ℂ}^{3}$ is very similar to a "clock" $ℤ}{12ℤ}\text{.}$

Use row reduction to find invertible matrices $L$ and $R$ (with entries in $ℂ\left[t\right]\text{)}$ such that $L\left(t-A\right)R$ is diagonal.

$A=\left(\begin{array}{cc}3& -7\\ 1& 5\end{array}\right)\text{.}$ Then $(t-A)= ( t-37 -1t-5 )$ and $( t-37 -1t-5 ) ⟶ ( -1t-5 t-37 ) ⟶ ( 1-(t-5) t-37 ) ⟶ ( 1t-5 07+(t-5)(t-3) ) ⟶ ( 1t-5 0t2-8t+22 ) ⟶ ( 10 0t2-8t+22 ) .$ ${t}^{2}-8t+22=\left(t-\lambda \right)\left(t-\mu \right)$ where ${μ,λ}= {8±64-442}= {8±202}= {4±5}$ and $L = (1-(t-3)01) (-1001) (0110)= (-(t-3)-110), R = (1-(t-5)01).$ So $L(t-A)R= L(t-37-1t-5)R= (100t2-8t+22)$ and $ℂ[t]⊕2 (t-A)ℂ[t]⊕2 ⥲ Lℂ[t]⊕2 L(t-A)ℂ[t]⊕2 = ℂ[t]⊕2 L(t-A)Rℂ[t]⊕2 (g1g2) ⟼ L(g1g2)$ and $ℂ[t]⊕2 (100t2-8t+22)ℂ[t]⊕2 = ℂ[t]ℂ[t]⊕ ℂ[t](t2-8t+22)ℂ[t]= ℂ[t](t2-8t+22)ℂ[t].$ $\frac{ℂ\left[t\right]}{\left({t}^{2}-8t+22\right)ℂ\left[t\right]}$ has basis $\left\{1,t\right\}$ with $t\text{-action}$ on this basis given by the matrix $\left(\begin{array}{cc}0& -22\\ 1& 8\end{array}\right)\text{.}$

## Notes and References

These are a typed copy of Lecture 36 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 28, 2011.