Genady

The following are my understandings related to the exercise in the previous post. A field of pn elements consists of roots of the polynomial xq-x, where q=pn. These are combined roots of all irreducible polynomials over field Fp which divide xq-x. These are irreducible polynomials of degrees m such that m | n. Each of them contributes m roots to the field. The number of elements in the field, pn, is sum of products m·Np(m), where Np(m) is number of irreducible polynomials of degree m over field Fp. This sum is for all m which divide n.

What conspiracy? What $7?

Moreover, the new issue CD for a month today gives 4.2% yield. It's $7 a month on $2000 deposit.

Ah, but your original explanation was different, i.e.,

Show that a field L with q = pn elements contains a field K with r = pm elements if and only if m | n. A field with pn elements has degree n over the prime field Fp, [L : Fp] = n. A field with pm elements has degree m over the prime field Fp, [K : Fp] = m. If L contains K, then [L : Fp] = [L : K] [K : Fp], i.e., n = [L : K] m. Thus, m | n. L consists of all roots of polynomial xq-x. K consists of all roots of polynomial xr-x. It can be proved by induction than if m | n, then all roots of the later are also roots of the former*. Thus, L contains K. QED * For example, if n = 2m then xq = (xr)r.

Write down a multiplication table for the elements of the finite field extension F2(α) where α2+α+1 = 0. All elements of the field F2(α) have a form a+bα, where a and b are 0 or 1. Thus, the elements are, 0, 1, α, and 1+α. The multiplication table is: 0 x anything = 0 1 x anything = the same anything α x α = 1+α α x (1+α) = 1 (1+α) x (1+α) = α The End.

I have had a paper related "glitch" with SSA, when they notified me that in their system, I am a non-citizen. They asked to send them the original naturalization certificate, which is almost impossible to replace, and promised to promptly send it back. It was in the beginning of 2020. A week after I mailed the certificate, the COVID shutdown started, and they could not access their mailroom. They have updated their system eventually, and I did get my certificate back... more than two years later.

This might be also a factor: SSA has assigned my application to their regional office in Santo Domingo, Dominican Republic, which probably operates on an "island time." 🙂

I will know the answer when I will see the payment amount.

It is fine if it is in fact so. However, I understood from SSA documents that the payment amount is determined by the month I've applied for it rather than by the first month I received it. Maybe I've misunderstood this point.

Thank you. I don't think I am eligible. I never worked in public sector or received a public pension.

I don't see, why. If I started receiving the money in November, I would invest it and start accumulating interest. Instead, the money is in "their" bank for months. I don't see what it has to do with the question. I know the answer to this question.

I could've calculated the total number of degree 5 irreducible polynomials without finding them as follows. The total number of degree 5 polynomials with constant term 1 is 16. From the list of irreducible polynomials of degrees 1 to 4 I find that there are 10 distinct combinations that make degree 5. These are reducible. Thus 16-10=6 irreducible polynomials left.

Show that if f(x) is an irreducible polynomial over K and K ⊂ L such that the degree of L over K and the degree of f(x) are relatively prime, then f(x) is irreducible over L. Let x0 be a root of f(x) and n be a degree of f(x). Since f(x) is irreducible over K, the degree of K(x0) over K is n. Let r be the degree of L over K. We know that n and r are relatively prime. Let's assume that f(x) is reducible over L. Then x0 is root of one of its factorizing polynomials. Call its degree, m, where m < n. Since this polynomial is irreducible over L, the degree of L(x0) over L is m. Since L includes K, L(x0) includes K(x0). Call the degree of L(x0) over K(x0), t. We have the degree of L(x0) over K equals nt, because of the inclusions K ⊂ K(x0) ⊂ L(x0), and equals rm, because of the inclusions K ⊂ L ⊂ L(x0). So, nt = rm. As n and r are relatively prime, n | m. But this contradicts that m < n. Thus, f(x) is irreducible over L. QED. Any doubts? 🙂

I heard this story ~50 years ago, but rather as a joke like this: Anyway, The Chicken Cannon | Snopes.com rates this story a legend.

Factorize x4+2 in F13[x]. I've found it irreducible. Is it correct and is there a shorter way? Here is my calculation: if it is reducible, then x4+2 = (x2+ax+b)(x2+cx+d)=x4+(a+c)x3+(ac+b+d)x2+(ad+bc)x+bd. Thus, a+c = 0, ac+b+d = 0, ad+bc = 0, bd = 2 (all mod 13). There are six (b,d) pairs such that bd = 2. They are (1,2), (3,5), (4,7), (6,9), (8,10), and (11,12). I will use the observation that for all of them d-b ≠ 0 and b+d ≠ 0. From a+c = 0 we get c=-a, and substituting in the other two equations, -a2+b+d = 0, a(d-b) = 0. Since d-b ≠ 0, a = 0. Thus b+d = 0, which is not so. Thus, we can't solve for the a, b, c, d. So, the polynomial is irreducible.

Factorize X7+1 in F2[X]. One root is 1. It makes X7+1 = (X+1)(X6+X5+X4+X3+X2+X+1). Checking against the list of irreducible polynomials of degrees 2 and 3 in the post at the top of this thread gives the factorization of the degree 6 polynomial above: (X3+X2+1)(X3+X+1).

Thanks a lot. I looked at my calculations. Turns out I've checked all three of them and made sloppy mistakes.

Factorize \(X^4+64\) in \(\mathbb Q [X]\). \(X^4+64=X^4+16X^2+64-16X^2=(X^2+8)^2-16X^2=(X^2+4X+8)(X^2-4X+8)\) Factorize \(X^4+1\) in \(\mathbb R [X]\). \(X^4+1=X^4+2X^2+1-2X^2=(X^2+1)^2-2X^2=(X^2-\sqrt 2 X +1)(X^2+\sqrt 2 X +1)\)

A much simpler proof that I've missed above: Since \(\frac p q\) is a solution, we have \(a_n(\frac p q)^n+a_{n-1}(\frac p q)^{n-1}+...+a_0=0\). Multiplying by \(q^n\) we get \(a_np^n+a_{n-1}p^{n-1}q+...+a_0q^n=0\). Thus, \(p|a_0q^n\). So, \(p|a_0\). And thus, \(q|a_np^n\). So, \(q|a_n\). QED

Could you write them down, please? I assume I'll get to them eventually. Thanks.

There are two (2) New Years in Russia, the New Year and the Old New Year: (Old New Year - Wikipedia)

Hmm... I've checked it and found it reducible: x5+x+1 = (x2+x+1)(x3+x2+1).

Show that if a rational number \(\frac p q\), where p, q are relatively prime integers, is a solution of an equation \(a_nX^n+...+a_0=0\) with all integer coefficients \(a_i\), then \(p|a_0\) and \(q|a_n\). Since \(\frac p q\) is a root of the polynomial on the left, \(a_nX^n+...+a_0=(X-\frac p q)(b_{n-1}X^{n-1}+...+b_0)\), where all coefficients \(b\) are rational. According to Gauss's Lemma, there exist rational numbers \(r,s\) such that \(rs=1\) and all the coefficients of the polynomials \(r(X-\frac p q)\) and \(s(b_{n-1}X^{n-1}+...+b_0)\) are integers. \(a_0=r \frac p q s b_0\). Since \(p|r \frac p q s b_0\), \(p|a_0\). \(r \frac p q\) is integer, thus \(q|r\). \(a_n=r s b_{n-1}\). Since \(q|r\), \(q|a_n\). QED

No, it does not. All the referred principles show that something exists. They don't show that nothingness does not exist.