Question about Euclidean algorithm

Minhyong Kim,
I am a student on your algebra 3 course:Maths 2201. Please could you email me an example of how to use Euclid’s theorem on two rings and how to use Bezout’s lemma on the same two rings, to help me with coursework sheet 3. Also how do you show
hcf(2,x+1)=1.

Many Thanks

Reply:

The point is to use the algorithm exactly the same way as for the integers. For example:

(x^4-1,x^3-1)=(x-1,x^3-1)

since

x^4-1=x(x^3-1)+x-1

But then

(x-1,x^3-1)=x-1

since (x^3-1)=(x-1)(x^2+x+1).

In this example, note that

x-1=(x^4-1)+(-x)(x^3-1)

giving an explicit form to Bezout’s lemma. In general, this would have to be iterated to relate the GCD back to the original pair.

As for a statement like

(2,x+1)=1

this follows from the definitions. Obviously, 2 (being a unit) divides (x+1) in Q[x] and no polynomial of degree 1 will divide both. So 2 is a divisor of maximal degree. But then, by convention, the GCD is supposed to be monic i.e., have leading coefficient 1 among those of maximal degree dividing both. See def. 2.2.31 in the notes.

In the case of constant polynomials, this means the GCD is 1. So in general, whenever you are doing the algorithm in a polynomial ring k[x] where k is a field, and you arrive at a situation like

(c, r(x))

where c is a non-zero constant, the GCD is 1. Incidentally, in this situation, a straightforward application of the procedure to relate back to the original pair will produce

c=af+bg

But then

1=(a/c)f+(b/c)g

MK

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