# Greatest common divisor of two polynomials

#### Peter

##### Well-known member
MHB Site Helper
I am working on Exercise 8 of Dummit and Foote Section 9.2 Exercise 8

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Determine the greatest common divisor of [TEX] a(x) = x^3 - 2 [/TEX] and [TEX] b(x) = x + 1 [/TEX] in [TEX] \mathbb{Q} [x] [/TEX]

and write it as a linear combination (in [TEX] \mathbb{Q} [x] [/TEX] ) of a(x) and b(x).

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In working on this I applied the Division Algorithm to a(x) and b(x) resulting in

[TEX] x^3 - 2 = (x^2 - x + 1) (x+ 1) + (-3) [/TEX]

then

[TEX] (x + 1) = (1/3 x + 1/3) + 0 [/TEX]

Last non-zero remainder is -3

Therefore, gcd is -3

BUT!

This does not seem to be correct because -3 does not divide either a(x) and b(x)

Peter

#### Opalg

##### MHB Oldtimer
Staff member
I am working on Exercise 8 of Dummit and Foote Section 9.2 Exercise 8

==================================================

Determine the greatest common divisor of [TEX] a(x) = x^3 - 2 [/TEX] and [TEX] b(x) = x + 1 [/TEX] in [TEX] \mathbb{Q} [x] [/TEX]

and write it as a linear combination (in [TEX] \mathbb{Q} [x] [/TEX] ) of a(x) and b(x).

==================================================

In working on this I applied the Division Algorithm to a(x) and b(x) resulting in

[TEX] x^3 - 2 = (x^2 - x + 1) (x+ 1) + (-3) [/TEX]

then

[TEX] (x + 1) = (1/3 x + 1/3) + 0 [/TEX]

Last non-zero remainder is -3

Therefore, gcd is -3

BUT!

This does not seem to be correct because -3 does not divide either a(x) and b(x)

$-3$ is a unit in $\mathbb{Q} [x]$, so is equivalent to $1$. You have shown that $$-\tfrac13(x^3-2) + \tfrac13(x^2-x+1)(x+1) = 1.$$ Thus $p(x)a(x)+q(x)b(x) = 1$, where $p(x) = -\frac13$ and $q(x) = \frac13(x^2-x+1)$. The polynomials $p(x)$ and $q(x)$ are both in $\mathbb{Q} [x]$.