Given the field extension `ZZ_2(alpha)` over the base field `ZZ_2` , and assuming that `alpha` is a root of the irreducible polynomial `m_{alpha}(x) = x^3 + x + 1 in ZZ_2[x]`. Let `beta = 1 + alpha`. We use the premises that:
`diamond beta in ZZ_2(alpha)`.
`diamond beta` is algebraic over `ZZ_2`.
The task is to find the irreducible polynomial (i. e. the minimal polynomial) for `beta = 1 + alpha in ZZ_2[x]`.

General Strategy for a Solution:
1. Use the Algebraic Premise: Since `beta` is algebraic over `ZZ_2`, we know a non-zero polynomial `p(x) in ZZ_2[x]` exists such that `p(beta) = 0`.
2. Relate `beta` and `alpha`: Use the definition `beta = alpha + 1` to express `alpha` in terms of `beta`. So , `alpha = beta - 1 = beta + 1` `(-1 equiv 1 mod 2)`.
3. Construct p(x): Substitute the expression for `alpha` into the known minimal polynomial equation for `alpha` , `m_alpha(alpha) = 0`.
4. Simplify and Factor: Simplify the resulting polynomial `p(beta)` using the arithmetic rules of `ZZ_2` (mod 2 , where 1 + 1 = 0). The minimal polynomial `m_beta(x)` will be the monic , irreducible factor of p(x) of the lowest degree.

Specific Theory Used:
`diamond` The Theorem: The premise that `bb{beta}` is algebraic over `bb{ZZ_2}` stems from the statement that every element of a finite extension `F(alpha)` is algebraic over F.
`diamond` Minimal Polynomial: The irreducible polynomial we seek , `m_beta (x)` , is the unique monic polynomial of the smallest degree in `ZZ_2[x]` such that `m_beta(beta) = 0`.
`diamond` Arithmetic in `bb{ZZ_2}`: The field `ZZ_2 = {0 , 1}` operates under modulo 2 arithmetic: 1 + 1 = 0 , -1 = 1 , and `(a + b)^p = a^p + b^p` when p is a prime (like p = 2 or p = 3) is often useful.

Do the exercise:
We use the assumption that `alpha` is a root of `color(blue)(m_alpha(x) = x^3 + x + 1) implies alpha^3 + alpha + 1 = 0`.
Step 1: Relate `alpha` and `beta`: `beta = alpha + 1 implies alpha = beta - 1 = beta + 1` (mod 2).
Step 2: Construct the polynomial `p(beta)`: Substitute `alpha = beta + 1` into the minimal polynomial equation for `alpha`: `color(blue)((beta + 1)^3 + (beta + 1) + 1 = 0)`.
Step 3: Simplify the Polynomial over `Z_2`:
1. Expand the cubic term: `(beta + 1)^3 = beta^3 + 3 beta^2 + 3 beta + 1`. Using `3 equiv 1` (mod 2) , or simply knowing that that all even binomial coefficients drop out in `ZZ_2` due to `2 equiv 0` :
`color(blue)((beta + 1)^3 = beta^3 + beta^2 + beta + 1)`.
2. Substituting back: `(beta + 1)^3 + (beta + 1) + 1 = 0 implies (beta^3 + beta^2 + beta + 1) + (beta + 1) + 1 = 0 implies`
`beta^3 + beta^2 + (beta + beta) + (1 + 1 + 1) = 0 implies color(blue)(beta^3 + beta^2 + 1)`.

Conclusion: The resulting irreducible polynomial for `beta = 1 + alpha in ZZ_2[x]` is `bb{m_{1 + alpha}(x) = x^3 + x^2 + 1}`.
This polynomial is irreducible over `ZZ_2`. You can check by testing x = 0 and x = 1. This shows that `alpha` and `alpha + 1` are roots of two different irreducible polynomials of the same degree 3. END