Comments for Bloomsbury Journal

Comment on Inverses in algebraic number fields by minhyong kim

minhyong kim — Thu, 31 Mar 2011 22:59:22 +0000

Sorry, I can’t answer questions like that.

Comment on Inverses in algebraic number fields by jack

jack — Thu, 31 Mar 2011 22:26:57 +0000

Will this kind of question comes up in the exam??? ie the Inverses

Comment on Beginning questions in number theory by minhyong kim

minhyong kim — Mon, 17 Jan 2011 10:15:01 +0000

For a given field extension of , it is true that there are many possible irreducible polynomials such that . However, it is definitely *not* true that all irreducible polynomials of the same degree define isomorphic field extension. Consider and and ask yourself what kind of field extensions of we get in the two cases. It is actually a delicate question to determine when two irreducible polynomials give isomorphic field extensions. We will discuss this in class quite a bit.

Comment on Beginning questions in number theory by Sandeep

Sandeep — Sun, 16 Jan 2011 09:13:11 +0000

If you take two different irreducible polynomials f and g in K[X], of the same degree d, and construct the fields K[X] / (f), and K[X] / (g), then is it correct that these will both be the “same” field? Because as vector spaces over K[X] they will both have the same dimension d. If they are the same, then does this mean that there is no unique irreducible polynomial f in K[X] which defines a field extension of K[X] of degree d?

Thanks in advance.

Sandeep.

Comment on Computing orders mod 23 by Tomjunlee

Tomjunlee — Sat, 26 Sep 2009 17:47:48 +0000

Not sure if this helps or is even relevant:

You can use a fast powering algorithm (number of operations reduces to logarithmic instead of linear) -

To calculate a^k (mod m):

(1) Form a table with three columns and first row: “k” “a” “1″;

(2) at each stage, form a new row as follows:

(a) the entry in column 1 is half the entry in column 1 of the previous row, ignoring fractions;

(b) the entry in column 2 is the square of the entry in column 2 of the previous row (modulo m);

(c) if the entry in column 1 of the previous row is even, the entry in column 3 is the same as the entry in column 3 of the previous row, otherwise it is the product (modulo m) of the entries oin column 2 and column 3 of the previous row.
—
(3) when the entry in column 1 is 0, the entry in column 3 of that row is the required power.

EXAMPLE: To calculate 2^11 (mod 23):

11 2 1
5 4 2
2 16 8
1 3 8
0 9 1

Therefore, 2^11 = 1 (mod 23).

Comment on Norms of elements and principality by minhyong kim

minhyong kim — Fri, 15 May 2009 08:59:02 +0000

`Exhaust other possibilities’ sounds exhausting. But recall that the equation

for an unknown element in our ring is not too difficult to analyze, at least in many examples. If there are no solutions at all, then a fortiori there are no solutions in our ideal. So the ideal would not be principal. However, if there is a solution, you would still need to check if there is a solution in the ideal. This can be somewhat trickier.

Comment on Norms of elements and principality by rory

rory — Thu, 14 May 2009 19:31:02 +0000

Thanks for the feedback. So, if I understand you correctly, given a situation where we are checking for principality of an ideal, but we don’t find a generator with our first presentation, our only option would be to exhaust all the other possibilities?

Comment on Norm of an ideal by Wai-Kae

Wai-Kae — Fri, 01 May 2009 10:51:36 +0000

I think I see it now,
The definition is

N(J) = |O_F / J |

So in the example,

Z[i]/(5,1+i) = {0}

=> N(J) = |{0}| = 1

I’m afraid I might have overseen this amidst the chaos… sorry..!..

Comment on Unique factorization by minhyong kim

minhyong kim — Wed, 08 Oct 2008 20:55:41 +0000

Yes, provided you could factor

Remember, it is not itself a prime number in general. It just has a prime number factor that doesn’t belong to the list. Given a number that’s not a prime, it is exactly the problem of finding a prime number factor that admits no efficient algorithm.

Recall that if you’re satisfied with an inefficient algorithm, then you can keep finding prime numbers by trying to divide numbers one by one.

Comment on Unique factorization by Kunal Patel

Kunal Patel — Wed, 08 Oct 2008 20:41:57 +0000

sir i was just wondering, you know in the proof of the theorem that there are infinitely many prime numbers, doesnt the line p1.p2.p3…..pn +1 provide a simple algorithm for finding prime numbers? albeit, requiring a lot of computing!
vikram mukarji