## Monthly Archives: October 2007

### Question on sheet 3, no. 1

Dear Sir,

I am currently stuck on part 3 of question 1 of the sheet3. Would you please give me a hint about that. Many Thanks!

Reply: Think carefully about 1.(ii) and combine it with the given fact that f(alpha)=0.

MK

### Question about F_5

I am currently stuck on question 2 of problem sheet 3. After changing the polynomials into their mod 5 forms, do I divide them to find the hcf as normal (without changing the remainders etc into their mod 5 forms) or after every division work out the mod 5 forms of the polynomials. For the Bezout’s lemma will I work out h and k and then convert them into their mod 5 forms right at the end. Sorry if that is a little confusing. Many Thanks.

One way is to divide normally, just remembering at each stage to set any multiple of 5 to zero. At the end of all your calculations, convert to elements of F_5[x]. For example, if your answer comes out to be

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

in F_5[x] this would become

x^3+x+2

since -4=1 and (1/3)=2 in F_5. For small calculations done by humans, this is probably the easiest process, at least from a psychological standpoint. However, you should be aware that for large calculations done by machines (or by humans used to F_5) converting at each stage to the F_5 form is much more efficient.

Feel free to ask again after trying this out.

MK

### 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

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

### Dear Students,

The intention of this blog is to facilitate communication with my current students. I expect many of the discussions to start from replies to email queries on specific mathematical issues. But if there are other general topics you would like me to comment on, please let me know using the submission form, by email, or by attaching a comment to a post here.

I will include occasionally some remarks on the topics that come up in class, so check back frequently.

MK