## Group Theory and Linear Algebra

Last updated: 20 September 2014

## Lecture 6: Greatest common divisors and Euclid's algorithm

### Number systems – $𝔽\left[t\right]$ polynomials

Let $𝔽$ be a field. $𝔽[t]= { a0+a1t+ a2t2+⋯ | ai∈ 𝔽,all but a finite number of ai equal 0 }$ with addition and product so that $(5+t+7t2)+ (3+2t+8t2+(-3)t3) =8+3t+15t2+4t3,$ $(1+2t2+t3) (0+0t+t2+2t3) = t2+2t3+4t4+ 8t5+t5+2t6 = t2+2t3+4t4+ 9t5+2t6,$ for example.

Let $d\in 𝔽\left[t\right]\text{.}$ The ideal generated by $d$, or the set of multiples of $d$, is $d𝔽[t]= {dp | p∈𝔽[t]}.$ For example, ${t}^{2}+2{t}^{3}+4{t}^{4}+9{t}^{5}+2{t}^{6}\in \left(1+2{t}^{2}+{t}^{3}\right)𝔽\left[t\right]\text{.}$

Let $a,d\in 𝔽\left[t\right]\text{.}$ The polynomial $d$ divides $a$, $d|a,$ if $a\in dℤ\text{.}$

$(1+2t2+t3)| (t2+2t3+4t4+9t5+2t6).$

Let $x,m\in 𝔽\left[t\right]\text{.}$ The greatest common divisor of $x$ and $m,$ $\text{gcd}\left(x,m\right)$ is a monic polynomial $d$ such that

 (a) $d|x$ and $d|m\text{,}$ (b) If $\ell \in {ℤ}_{>0}$ and $\ell |x$ and $\ell |m$ then $\ell |d\text{.}$

Let $p\in 𝔽\left[t\right],$ $p={p}_{0}+{p}_{1}t+{p}_{2}{t}^{2}+\cdots$ with $p\ne 0\text{.}$ The degree of $p$ is $N\in {ℤ}_{>0}$ such that ${p}_{N}\ne 0$ and if $k\in {ℤ}_{>0}$ and $k>N$ then ${p}_{k}=0\text{.}$

A polynomial $p={p}_{0}+{p}_{1}t+{p}_{2}{t}^{2}+\cdots$ is monic if ${p}_{N}=1,$ where $N=\text{deg}\left(p\right)\text{.}$

(Euclid's algorithm) Let $a,b\in 𝔽\left[t\right]\text{.}$ There exist $q,r\in 𝔽\left[t\right]$ such that

 (a) $a=bq+r,$ (b) Either $r=0$ or $\text{deg}\left(r\right)<\text{deg}\left(b\right)\text{.}$

Let $x,m\in 𝔽\left[t\right]\text{.}$

 (a) There exists a monic polynomial $\ell$ such that $ℓ𝔽[t]=x 𝔽[t]+m𝔽[t].$ (b) Let $d=\text{gcd}\left(x,m\right)\text{.}$ Then $d=ℓ.$

Find $gcd ( (x4-3x3+3x2-3x+2), (x3-10x2+23x-14) ) .$ $)x4+7 x3-10x2+23x-14 )x4-3x3+3x2-3x+2+1000 ‾ ) x4-10x3+ 23x2-14x )x4- 7x3-20x2+ 11x+2+10 ‾ )x4- 7x3-70x2+161x-98 )x4- 7x3-50x2-150x+100 ‾$ So $(x4+3x3+3x2-3x+2) = (x3-10x2+23x-14) (x+7)+50(x2-3x+2), x3-10x2+23x-14 = (x2-3x+2)(x-7).$ So $gcd(x4-3x3+3x2-3x+2,x3-10x2+23x-14) =x2-3x+2$ and $x2-3x+2=150 (x4-3x3+3x2-3x+2)+ (-150x-750) (x3-10x2+23x-14).$

## Notes and References

These are a typed copy of Lecture 6 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 5, 2011.