Let f(x) = x² − 2
It is in the ring of polynomials with integer coefficients.
Eisenstein's criterion proves a polynomial with integer coefficients is irreducible over Q if there exists a prime p such that: * p divides every coefficient except the leading one, * p does not divide the leading coefficient, * And p² does not divide the constant
Application to f(x) = x² − 2, coefficients (1, 0, −2): take p = 2. Then 2 divides the non-leading coefficients 0 and −2, 2 does not divide the leading coefficient 1, and 2² = 4 does not divide −2. so f is irreducible over Q.
Now consider the field of rational numbers adjoined with √2: Let K = Q(√2). Since √2 is a root of f and f is irreducible over Q, f is the minimal polynomial of √2 over Q. So the degree of the field extension K of Q is 2 ([K:Q]=2, the dimension of K as a vector space with scalars in Q).
If √2 were a member of Q, then K = Q, and [K : Q] = 1, Hence √2 is not in Q. The square root of 2 is irrational.
Furthermore, The Galois group of K over Q is isomorphic to integers mod 2 with the elements being the automorphisms: identity and swapping of √2 and −√2. You can use this to the same effect, but in reality the Galois Theory can't rigorously work in a proof because it ends up being circular since Galois is built on knowing facts about arithmetic and number fields.