Genady

Is the extension K ⊆ L normal, where K = Q, L = Q(21/3)? 21/3 is zero of polynomial x3-2, which is irreducible in Q. Other two zeroes of this polynomial are complex and thus ∉ L. Hence, this extension is not normal. Is such extension normal for K = Q(21/2), L = Q(21/4)? 21/4 is zero of polynomial x4-2, which is reducible in K, i.e., x4-2 = (x2 - 21/2)(x2 + 21/2). The first factor has both zeroes in L, while the second does not have any zeroes in L. Hence, this extension is normal.

The same meds as everyone else. You are right about the long-term treatment, but the question was,

We don't know. It's many years now she does not work.

May I offer anecdotal evidence? My wife used to be a psychiatric nurse in NY hospital in Greenwich Village, Manhattan, 30 years ago. You can imagine that they had a lot of emergency homeless patients there. They treated all of them well, in practice.

Let S ⊆ M ⊆ L be field extensions. 1. Let S ⊆ M and M ⊆ L be normal extensions. Is S ⊆ L normal? Not necessarily. Here is an example of it being "abnormal." Let S = Q, M = Q(21/2), L = Q(21/4). Polynomial x2-2 is irreducible over S and has both of its zeroes in M. So, S ⊆ M is normal. Polynomial x2-21/2 is irreducible over M and has both of its zeroes in L. So, M ⊆ L is normal. However, polynomial x4-2 is irreducible over S but has only two of its four zeroes in L. Thus, S ⊆ L is not normal. 2. Let S ⊆ L be normal. Is M ⊆ L normal? Yes. Any polynomial irreducible over M is irreducible over S. If this polynomial has one zero in L it has all zeroes in L, because S ⊆ L is normal. Since it has all its zeros in L, M ⊆ L is normal, too. 3. Let S ⊆ L be normal. Is S ⊆ M normal? Not necessarily. Let S = Q, M = Q(21/4), L = Q(21/4, i). Polynomial x4-2 is irreducible over S and has all its zeros in L. So, S ⊆ L is normal. However, it has only two of its four zeroes in M. Thus, S ⊆ M is not normal.

There is no need in referendums anymore. The data are there, in social media, forums, TV, papers, etc. Everything people talk and write about. All the answers. The tool to collect and summarize these data is also there. It is called, AI. /sarcasm/

The last formula in the previous post can be used to calculate the numbers of irreducible polynomials without listing them. For example, for p = 2 we get: degree 1 polynomials, N2(1) = 21/1 = 2, degree 2 polynomials, N2(2) = (22 - 1xN2(1))/2 = 1, degree 3 polynomials, N2(3) = (23 - 1xN2(1))/3 = 2, degree 4 polynomials, N2(4) = (24 - 1xN2(1) - 2xN2(2))/4 = 3, degree 5 polynomials, N2(5) = (25 - 1xN2(1))/5 = 6. These numbers can be compared with the finding in this earlier exercise:

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.