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

please help me! Thanks

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.