Make a list of all irreducible polynomials of degrees 1 to 5 over F2

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SUMMARY

The discussion centers on identifying all irreducible polynomials of degrees 1 to 5 over the finite field F2, specifically focusing on the polynomial X^5 + X^4 + X^2 + X + 1. This polynomial is confirmed to be irreducible, suggesting that it may have been overlooked in the textbook's solutions. Additional irreducible polynomials identified include X^5 + X^4 + X^3 + X + 1, X^5 + X^3 + X^2 + X + 1, and X^5 + X^4 + X^3 + X^2 + 1. The discussion highlights the complexities of checking irreducibility, particularly in the context of algebraic geometry.

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Hill
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TL;DR
F2 = (0, 1)
To this exercise,
1735996974647.png

the textbook's solution is
1735997016225.png

I didn't follow their method and have found another degree 5 solution, ##X^5+X^4+X^2+X+1##.
Is it wrong or the book has omitted this solution?
 
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Hill said:
TL;DR Summary: F2 = (0, 1)

To this exercise,
View attachment 355323
the textbook's solution is
View attachment 355324
I didn't follow their method and have found another degree 5 solution, ##X^5+X^4+X^2+X+1##.
Is it wrong or the book has omitted this solution?
It is irreducible. I guess they missed it.
 
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martinbn said:
It is irreducible. I guess they missed it.
As well as
##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##
(?)
 
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Hill said:
As well as
##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##
(?)
Yes.
There are six of them.
 
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This seems a little easier than the book's hints make it. A poly. over Z/2 is divisible by X or X+1 iff it either ends in X or has an even number of terms, hence the irreducible cubics end in 1 and have an odd number of terms, i.e. X^3+X^2 + 1, and X^3+X+1. A reducible quartic or quintic either ends in X or has an even number of terms or is divisible by X^2+X+1....It is still somewhat tedious, but maybe not as long as multiplying together all those factors to compute all reducible examples. It is interesting that after all these years I still have trouble checking irreducibility. I guess it is just inherently difficult. Such problems have been challenging for years in algebraic geometry: Severi's conjecture, irreducibility of moduli of curves, components of Hilbert schemes,....
 

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