Genady Posted January 3 Posted January 3 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?
Genady Posted January 3 Author Posted January 3 On 1/3/2025 at 5:42 PM, studiot said: x4 + x2 + 1 = (x2 - x + 1) (x2 + x + 1) Expand Yes. Thank you!
studiot Posted January 3 Posted January 3 On 1/3/2025 at 5:50 PM, Genady said: Yes. Thank you! Expand Thanks Interesting that your field now allows (0, 1). 😀
Genady Posted January 3 Author Posted January 3 On 1/3/2025 at 6:54 PM, studiot said: Thanks Interesting that your field now allows (0, 1). 😀 Expand Yes. It is the field (0, 1), not a set {0, 1} as a subset of C.
Genady Posted January 3 Author Posted January 3 (edited) 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 Edited January 3 by Genady
KJW Posted January 3 Posted January 3 (edited) On 1/3/2025 at 5:11 PM, Genady said: x5+x2+1, x5+x3+1, x5+x4+x2+x+1. Did I miss any? Expand x^5 + x + 1 In general, the root of x^5 + x + a is called a Bring radical of a, denoted BR(a). Bring radicals extend the notion of radicals with regards to the Abel-Ruffini theorem that states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. By extending the notion of radicals, quintic equations that have no solution in ordinary radicals have solutions in Bring radicals. Note that asymptotically, \text{BR}(a) \sim -a^{1/5} for large a. Edited January 3 by KJW
Genady Posted January 3 Author Posted January 3 On 1/3/2025 at 10:39 PM, KJW said: x5+x+1 Expand Hmm... I've checked it and found it reducible: x5+x+1 = (x2+x+1)(x3+x2+1).
KJW Posted January 3 Posted January 3 On 1/3/2025 at 10:54 PM, Genady said: Hmm... I've checked it and found it reducible: x5+x+1 = (x2+x+1)(x3+x2+1). Expand Perhaps I misinterpreted your original question concerning the field (0, 1). Anyway, I was focusing on the notion of Bring radicals because I thought they were interesting in terms of solving general quintic (and perhaps higher) equations.
uncool Posted January 4 Posted January 4 On 1/3/2025 at 5:11 PM, Genady said: 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? Expand There are a couple ways to check that may be beyond your current level; how much background do you have? Based on those ways, I'm pretty sure you've missed 3 irreducible degree-5 polynomials.
Genady Posted January 4 Author Posted January 4 On 1/4/2025 at 1:00 AM, uncool said: I'm pretty sure you've missed 3 irreducible degree-5 polynomials. Expand Could you write them down, please? On 1/3/2025 at 11:29 PM, KJW said: Perhaps I misinterpreted your original question concerning the field (0, 1). Anyway, I was focusing on the notion of Bring radicals because I thought they were interesting in terms of solving general quintic (and perhaps higher) equations. Expand I assume I'll get to them eventually. Thanks.
Genady Posted January 4 Author Posted January 4 (edited) On 1/3/2025 at 9:58 PM, Genady said: Show that if a rational number pq , where p, q are relatively prime integers, is a solution of an equation anXn+...+a0=0 with all integer coefficients ai , then p|a0 and q|an . Since pq is a root of the polynomial on the left, anXn+...+a0=(X−pq)(bn−1Xn−1+...+b0) , 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−pq) and s(bn−1Xn−1+...+b0) are integers. a0=rpqsb0 . Since p|rpqsb0 , p|a0 . rpq is integer, thus q|r . an=rsbn−1 . Since q|r , q|an . QED Expand 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 Edited January 4 by Genady
Genady Posted January 4 Author Posted January 4 (edited) 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) Edited January 4 by Genady
uncool Posted January 4 Posted January 4 On 1/4/2025 at 1:21 AM, Genady said: Could you write them down, please? Expand At the time, I didn't know what the polynomials were; the method I used told me the total number of degree-5 irreducible polynomials (6), not what they were. However, some basic manipulation gives me: x^5 + x^4 + x^3 + x + 1, x^5 + x^3 + x^2 + x + 1, x^5 + x^4 + x^3 + x^2 + 1 1
Genady Posted January 4 Author Posted January 4 On 1/4/2025 at 1:44 PM, uncool said: At the time, I didn't know what the polynomials were; the method I used told me the total number of degree-5 irreducible polynomials (6), not what they were. However, some basic manipulation gives me: x5+x4+x3+x+1,x5+x3+x2+x+1,x5+x4+x3+x2+1 Expand Thanks a lot. I looked at my calculations. Turns out I've checked all three of them and made sloppy mistakes.
Genady Posted January 4 Author Posted January 4 (edited) 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). Edited January 4 by Genady
Genady Posted January 4 Author Posted January 4 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.
Genady Posted January 5 Author Posted January 5 On 1/4/2025 at 1:44 PM, uncool said: At the time, I didn't know what the polynomials were; the method I used told me the total number of degree-5 irreducible polynomials (6), not what they were. However, some basic manipulation gives me: x5+x4+x3+x+1,x5+x3+x2+x+1,x5+x4+x3+x2+1 Expand 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.
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