# gcd of 2 polynomials over one Field included in another

I am trying to find the gcd of those 2 polynomials

$x^5-2x^2+1$ and $x^3+3x^2-x-3$

In the book I use, there is a corollary just before this exercice saying that

Corollary : Let $F$ $\subset$ $E$ be fields, and let $f(x),g(x)$ $\in$ F[$x$] $\subset$ E[x]. Then the gcd of $f$ and $g$ computed in $F[x]$ is the same as the gcd of f and g computed in $E[x]$.

My plan is then to find the gcd of those 2 polynomials in an "easy" fields, i.e $\mathbb{Z}_2[x]$

So, working in $\mathbb{Z}_2[x]$, our 2 polynomials become $x^5+1$ and $x^3+x^2-x-1$, is this true to start with ?

When I apply the euclidian algorithm, I find the quotient $q(x) = x^2-x+2$ and the remainder $r(x) = -2x^2+x+3$, i.e in $\mathbb{Z}_2[x]$, $q(x)=x^2-x$ and $r(x)=x+1$.

Applying again the algorithm i.e, euclidian division of $p(x)= x^3+x^2-x-1$ by $r(x)=x+1$, I find $r'(x) = 0$, $q'(x)=x^2-1$.

As the remainder is $0$, then the gdc of the 2 polynomials in $\mathbb{Z}_2[x]$ should be the former remainder, i.e $r(x)=x+1$ but it does not work in $\mathbb{Z}[x]$ ! Take $x=5$, the first polynomial gives 3076, the second one gives 192. The gdc should then be $x+1=5+1=6$ but $6$ does not divide $3076$.

In $\mathbb{Z}_2[x]$, the first polynomial gives $x^5+1=3126$, the second one $x^3+x^2-x-1=144$ Here, 6 indeed divides those 2 numbers, why does it not work in $\mathbb{Z}[x]$ ?

Which is not what is expected from the corollary.

Where did I go wrong ?

I tried to factorise the polynomial, they have a $1$ as a common root, so could $x-1$ be their gcd ? It is a common divisor but I do not know how to prove it is the greatest one.

Thanks very much for any help you can provide ! :D

• $\mathbb{Z}$ is not a field, and $\mathbb{Z}/2\mathbb{Z}$ is not even a subring of $\mathbb{Z}$. – Tobias Kildetoft Oct 1 '13 at 11:00
• Could you five me two fields I could use to solve this then ? Thanks for your answer ! – ALM Oct 1 '13 at 11:01
• You need to also use that monic polynomials factor the same over $\mathbb{Z}$ and over $\mathbb{Q}$, which will then allow you to use the corollary with for example $\mathbb{Q}$ and $\mathbb{C}$ (assuming you can factor them completely). – Tobias Kildetoft Oct 1 '13 at 11:04
• Do polynomial division by that root and see that there are no further common roots over $\mathbb{C}$ (you can easily find the roots of one of them and check that they are not roots of the other). – Tobias Kildetoft Oct 1 '13 at 11:13
• Yes, that is a correct argument, since you use the corollary to calculate the gcd over $\mathbb{C}$. – Tobias Kildetoft Oct 1 '13 at 11:27

## 1 Answer

We factorize first the 2 given polynomials such that :

$x^5−2x^2+1=(x−1)(x^4+x^3+x^2−x−1)$ and $x^3+3x^2−x−3=(x−1)(x^2+4x+3)$

They have the common root $1$. The others roots of the second polynomials are {-3,-1} but they are not roots of the first one, i.e $x-1$ is the only (then the greatest) common divisor of those 2 polynomials.

Keep in mind we use the corollary given in the original post to justify this argument.

Thanks for your help