*(From I.S.I. M.Math Subjective Sample Paper 2013)*

Let k be a field and k [x, y] denote the polynomial ring in the two variables x and y with coefficient from k . Prove that for any \(\mathbf{a, b \in k}\) the ideal generated by the linear polynomials x- a and y-b is a maximal ideal of k [x, y].

**Discussion:**

Suppose I = <x-a, y-b> is not the maximal ideal. Then there exist an ideal I’ such that \(\mathbf{I \subset I’ \subseteq k[x,y] }\). We show that I’ = k[x,y]

Suppose I’ contains I properly. There there exists an element P(x, y) (that is a polynomial in x, y with coefficients from k), which is not in I but in I’.

Say \(\mathbf{P(x,y) = \sum A_{m, n} x^m y^n ; A_{m, n} \in k}\). Since k[x,y] is ring of polynomials over k which is a field (hence a unique factorization domain), hence there exists unique Q(x,y) such that \(\mathbf{P(x,y) = Q(x,y) (x-a) + G(y) }\) where G(y) is a polynomial in y over k. As x – a is linear it will not leave any power of x in the remainder expression.

Again G(y) = R(y) (y-b) + T where T is in k. As (y-b) is linear the remainder won’t have any power of y.

Therefore \(\mathbf{P(x, y) = Q(x, y) (x-a) + R(y) (y – b) + T in I’ \implies T = P(x,y) – ( Q(x, y) (x-a) + R(y) (y – b)) \in I’}\)

But T is a constant polynomial in k[x,y]. Hence \(\mathbf{T \times T^{-1} = 1 \in I’}\) . Since I’ contains multiplicative identity element of field k it will absorb all polynomials from k[x,y], implying I’=k[x,y]. **Proved**