1. Let \(K\) be a field and let \(R=K[x]\) be the set of all polynomials in one variable over \(K\). So an element of \(R\) is a finite formal sum \(\sum_{i=0}^{d} a_{i} x^{i}\) with \(a_{i} ∈ K\).
    If \(f=\sum_{i} a_{i} x^{i}\) and \(g=\sum_{j} b_{j} x^{j}\) then define \(f+g\) and \(f × g\) in the obvious way: \(f+g=\) \(\sum_{i}\left(a_{i}+b_{i}\right) x^{i}\) and \(f g=\sum_{k} c_{k} x^{k}\) with \(c_{k}=\sum_{i+j=k} a_{i} b_{j}\).
    Prove that \(R\) becomes a commutative ring with a 1 with these definitions of + and \(×\).
  2. Let's goof around in \(ℚ[x]\).
  3. Prove that if \(f, g ∈ K[x]\) and at least one is non-zero, and if \(s, t\) are both gcd's of \(f\) and \(g\), then \(s=λ t\) for some \(λ ∈ K^{×}\).
  4. Factor the following polynomials in \(ℚ[x]\) into irreducible ones, giving proofs that your factors really are irreducible.