Why the polynomial of GCM is primitive?

Ievgeni

12/20/23, 10:30 AM

I'm interested on the polynomial used in GCM-mode : $X^{128}+X^7+X^2+X+1$

This polynomial is Primitive (in $\mathbb{F}_2$). What is the interest of choosing a primitive polynomial and not a simple irreducible polynomial? Is it a coincidence?

376

2 + 0

ctr

gcm

modular-arithmetic

Score:5

Crypto

Daniel S

12/20/23, 11:15 AM

I think that it is more relevant that this is the lexicographically-first degree 128 polynomial that is irreducible. This follows the example of the AES polynomial $X^8+X^4+X^3+X+1$ which is also the lexicographically-first irreducible of degree 8 (though not primitive). The primitivity is, I think, coincidental.

In the degree 128 case, the lexicographic choice leads to more efficient reduction processes, but probably just the lexicographically-first=nothing-up-my-sleeve is the principal motivation.

+ 1

kodlu

12/20/23, 1:33 PM

could the ability to generate an arbitrary 'phase' bitstring via a primitive LFSR be relevant?

Score:5

Crypto

Marc Ilunga

12/20/23, 11:29 AM

The choice of the polynomial is explained in the GCM specification. Being primitive didn't really play a role, but the designers were interested in a low-weight irreducible polynomial that in turns allow efficient implementation. The GCM polynomial is found in the table of low weight polynomials in this document (searching for 128,7,2,1).

+ 0

Elon Musk

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