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I have a question about $z$ in Classic Mceliece Algorithm specification. enter image description here

I have no idea about this $z$! In parameter set kem/mceliece348864, Field polynomial $f(z) = z^{12} + z^3 + 1$. is this $z$ in field polynomial same as the $z$ in pic? If this is right, the value of $z$ in the pic for kem/mceliece348864 is $(z^1, z^2, z^3, \dots, z^{11}) = (0, 0, 1, 0, \dots, 0)$?

please help me! Thanks

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    $\begingroup$ Could you give a link of the paper? $\endgroup$
    – Ievgeni
    Mar 9, 2022 at 12:10

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Yes, $z$ is the root of the polynomial used to construct the field (in the case of mceliece 348864 this field is $\mathbb F_{2^{12}}$ and the polynomial is as quoted). I'm not sure to which pic you refer, but if we choose to represent elements of $\mathbb F_{2^{12}}$ as 12-tuples of bits corresponding to the coefficients of the monomial basis elements $(1,z,z^2,z^3,\ldots,z^{11})$ then we would represent 1 as $(1,0,0,0,\ldots, 0)$; $z$ as $(0,1,0,0,\ldots,0)$ and so on. This means for example that in this case the element $\beta_0$ would be represented as $(d_0,d_1,d_2,d_3,\ldots,d_{11})$; $\beta_1$ would be represented as $(d_{\sigma_1},d_{\sigma_1+1},d_{\sigma_1+2},d_{\sigma_1+3},\ldots,d_{\sigma_1+11})$ and so on.

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  • $\begingroup$ Thanks! I understood!! I have one more quesition. I'm studying Classic McEliece Round 3 submission documentation. In page 14 of document, there is Irreducible-polynomial generation algorithm. But in page 19, irreducible polynomial $y^{64} + y^3 + y+z$ is defined for $F_{q}[y]$. this polynomial is g for key genearation?? $\endgroup$
    – Min
    Mar 14, 2022 at 11:39
  • $\begingroup$ No! The polynomial $g$ is user specific and must remain secret. The polynomial that you quote takes the role of $F(y)$ in line 2 of section 2.4.1 as reproduced in your question.. $\endgroup$
    – Daniel S
    Mar 15, 2022 at 14:06

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