Binius 1

Building extension filed:

We construct the field:\(\mathbb{F}_2[X_0]/(X_0^2 + X_0 + 1)\)

1. Start with \(\mathbb{F}_2[X_0]\): all polynomials in variable \(X_0\)with coefficients in \(\mathbb{F}_2\)​ (i.e., 0 or 1).
2. Then we take equivalence classes modulo \(X_0^2 + X_0 + 1\).
   That means two polynomials are considered the same if they differ by a multiple of \(X_0^2 + X_0 + 1\).
3. Because we’re reducing by a degree-2 polynomial, every element in the quotient can be represented as a polynomial of degree < 2, i.e.: \(a + bX_0, \quad \text{where } a,b \in \mathbb{F}_2\)
4. So we have only 4 distinct elements: \(\{0, 1, X_0, 1 + X_0\}\)

two steps:
1. polynomials ring: a polynomial with coefficients from F2, mod the irreducible polynomial, the result is a polynomial of format a+bx0,  x0 here is the indeterminate. as you see the notation F2[x0]/... "[x0]" means it is a variable.
2. use X0 to form F4, and the elements look like the same as "reminder" polynomials, but x0 now is not a variable, it is a constant. it is the root of the irreducible polynomial. 

The polynomial\(X_0^2 + X_0 + 1\) is irreducible over \(\mathbb{F}_2\), which means:

• It has no roots in \(\mathbb{F}_2\)​ — that is, no value \(x \in \mathbb{F}_2\)​ satisfies \(x^2 + x + 1 = 0\)
• So we introduce a new symbol \(X_0\)​ to “pretend” to be a root
• This symbolic root lives in the extension field \(\mathbb{F}_4\)​, not in \(\mathbb{F}_2\)​

🔁 Analogy:

It’s just like how i (the imaginary unit) isn’t a real number, but we construct the field \(\mathbb{C} = \mathbb{R}[X]/(X^2 + 1)\), and then say: \(X^2 + 1 = 0 \quad \Rightarrow \quad X = i\)

Here, XXX is a formal symbol, and once we mod out, it behaves like a concrete number — even though it’s not in \(\mathbb{R}\).

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✅ Final Clarification

❗️”Is \(X_0\) a field element?”

• Before reduction: No, it’s a formal variable
• After reduction (in \(\mathbb{F}_2[X_0]/(X_0^2 + X_0 + 1)\):
  ✔ Yes, \(X_0X\) becomes a field element in \(\mathbb{F}_4\)​, because we defined our field to treat it as such
  ✔ It represents a root of the irreducible polynomial — even though no such root existed in \(\mathbb{F}_2\)