Polynomials and Irreducibility exercises

[Genady]

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?

[studiot]

x⁴ + x^{2 +} 1  =  (x² - x + 1) (x² + x + 1)

[Genady]

7 minutes ago, studiot said:

x⁴ + x^{2 +} 1  =  (x² - x + 1) (x² + x + 1)

Yes. Thank you!

[studiot]

Just now, Genady said:

Yes. Thank you!

Thanks

 

Interesting that your field now allows (0, 1).

 

😀

 

[Genady]

43 minutes ago, studiot said:

Thanks

 

Interesting that your field now allows (0, 1).

 

😀

 

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

[Genady]

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 Friday at 10:02 PM by Genady

[KJW]

5 hours ago, Genady said:

x⁵+x²+1, x⁵+x³+1, x⁵+x⁴+x²+x+1.

Did I miss any?

[math]x^5 + x + 1[/math]

In general, the root of [math]x^5 + x + a[/math] 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, [math]\text{BR}(a) \sim -a^{1/5}[/math] for large a.

 

Edited Friday at 10:41 PM by KJW

[Genady]

13 minutes ago, KJW said:

x5+x+1

Hmm... I've checked it and found it reducible:

x⁵+x+1 = (x²+x+1)(x³+x²+1).

[KJW]

3 minutes ago, Genady said:

Hmm... I've checked it and found it reducible:

x⁵+x+1 = (x²+x+1)(x³+x²+1).

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]

7 hours ago, 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?

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]

19 minutes ago, uncool said:

I'm pretty sure you've missed 3 irreducible degree-5 polynomials.

 

Could you write them down, please?

1 hour ago, 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.

 

I assume I'll get to them eventually. Thanks.

[Genady]

9 hours ago, 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

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 Saturday at 07:55 AM by Genady

[Genady]

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 Saturday at 12:17 PM by Genady

[uncool]

12 hours ago, Genady said:

Could you write them down, please?

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:

[math]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[/math]

 

[Genady]

20 minutes ago, 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

 

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

[Genady]

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 Saturday at 04:04 PM by Genady

[Genady]

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]

On 1/4/2025 at 9:44 AM, 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

 

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.