Genady

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.

You're welcome 🙂.

Yes. It is the field (0, 1), not a set {0, 1} as a subset of C.

Yes. Thank you!

Make a list of all irreducible polynomials of degrees 1 to 5 over the field (0, 1). In the order of their degrees, this is my list: x, x+1; x2+x+1; x3+x+1, x3+x2+1; x4+x+1, x4+x2+1, x4+x3+1, x4+x3+x2+x+1; x5+x2+1, x5+x3+1, x5+x4+x2+x+1. Did I miss any? Is any of the above reducible?

This is the last exercise on this topic. Let \(K \subseteq L\) be a field extension and let \(M_1,M_2\) be two fields containing \(K\) and contained in \(L\). Show that \(M_1M_2\) consists of all quotients of finite sums \(\sum a_ib_i\) where \(a_i \in M_1, b_i \in M_2\). Any such quotient is constructed from elements of \(M_1, M_2\) by addition, multiplication, and inverses and thus it \(\in M_1M_2\). OTOH, a set of all such quotients constitutes a field that contains elements of \(M_1, M_2\) and thus must include \(M_1M_2\) because the latter is a smallest such field. Thus, they are equal.

This is a good one. Nicer than what I had in mind.

I think that the proposed principle is wrong and absolute nothingness is possible.

Is there any reasoning behind this question? The number you propose is close but does not seem to be exact.

Express \(\sqrt 5\) by rational numbers and a number \((\sqrt 5 + \sqrt 7)\) using addition, subtraction, multiplication and division.

Let \(K(\sqrt a)\) and \(K(\sqrt b)\), where \(a, b \in K; ab \neq 0\), be two field extensions of \(K\). Show that \(K(\sqrt a) = K(\sqrt b)\) if and only if \(ab\) is a square in \(K\) (that is, there is a \(c \in K\) such that \(ab = c^2\)). 1. If \(K(\sqrt a) = K(\sqrt b)\), then \(\sqrt a=r+s \sqrt b\) for some \(r, s \in K\). Then, \(\sqrt a - s \sqrt b =r\). Squaring it, \(a + s^2 b - 2s \sqrt {ab} = r^2\). Or, \(2 s \sqrt {ab} = a + s^2 b - r^2\). Since the RHS in the last equation is in \(K\), so must be its LHS. So, \(\sqrt {ab} \in K\) and thus, \(ab\) is a square in \(K\). 2. If \(ab = c^2\), then \(\sqrt {ab} = c, \sqrt a = c (\sqrt b)^{-1}\) and thus, \(K(\sqrt a) = K(\sqrt b)\). QED

This exercise generalizes the case of \( i^2=−1\). Let \( L \supset K \) be a field extension and let \(\alpha \in L \setminus K, \alpha^2 \in K\). Show that \[K(\alpha)=\{a+b\alpha; a,b \in K\}.\] Let's denote \(\alpha^2=r \in K\). Any polynomial \(a+b \alpha + c \alpha^2 +d \alpha^3 +... = a+b \alpha + c r +d r \alpha +... =(a+cr +...) + (b+dr+...)\alpha = s+t\alpha\), where \(s,t \in K\). Inverse of a polynomial, \((a+b\alpha)^{-1}=(a-b\alpha)(a^2-b^2r)^{-1} = a(a^2-b^2r)^{-1} - b(a^2-b^2r)^{-1}\alpha= s+t\alpha\), where \(s,t \in K\). QED

Show that if the smallest subfield K of a field L has an order n then na=0 for all a in L. The smallest subfield with an order n is {0, e, 2e, ... , (n-1)e} with ne=0. For any integer k, n(ke) = k(ne) = k(0) = 0. For x in L\K, nx = n(ex/e) = (ne)x/e = 0(x/e) = 0 QED

This was a challenging exercise. Give an example of an infinite field with a finite subfield. The simplest example I could come up with was a field of rational functions, \[\frac {a_n X^n+a_{n-1} X^{n-1}+...+a_0}{b_m X^m+b_{m-1} X^{m-1}+...+b_0}\] where all the coefficients a and b are from the field (0,1) as in the second post on the top of this thread,

They do, with \(b=0\) and a rational \(a\).