# What is $z$ in specification of Classic Mceliece?

I have a question about $$z$$ in Classic Mceliece Algorithm specification.

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 $$(z^1, z^2, z^3, \dots, z^{11}) = (0, 0, 1, 0, \dots, 0)$$?

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.
• 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??
• 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.. Mar 15, 2022 at 14:06