By Jan Nekovar

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15) Proposition. Let K → L be a field extension. (1) L/K is a finite extension =⇒ L/K is algebraic and of finite type. (2) If L = K(α1 , . . , αn ), where each αi is algebraic over K, then the extension L/K is finite, hence algebraic. (3) The implication “=⇒” in (1) is an equivalence. (4) K = {α ∈ L | α is algebraic over K} is a subfield of L. Proof. (1) If α1 , . . , αd is a basis of L/K (d = [L : K] < ∞), then L = K(α1 , . . , αd ); thus L/K is a field extension of finite type. If β ∈ L, then the d + 1 elements 1, β, .

Dr ) depend only on A: r is the rank of A (considered as a matrix with entries in the fraction field of R) and d1 · · · dk is the greatest common divisor of all k × k minors of A. Proof. (i) We can assume that A = 0. By induction, it is enough to transform A into d1 0 0 d1 B B ∈ Mn−1,m−1 (R) , ( ) by applying row operations A → gA, g ∈ GLn (R) (resp. column operations A → Ah, h ∈ GLm (R)). In particular, we can permute the rows (resp. the columns). We use the following observation: if a column C (resp.

10) Exercise. If R is a noetherian ring, so is the power series ring R[[X]]. 11) Proposition. In an noetherian ring R, every ideal I contains a suitable product P1 · · · Pr (r ≥ 0) of prime ideals (in particular, the zero ideal (0) is a product of prime ideals). Proof. Let S be the set of all ideals of R which do not contain any product of prime ideals. If S is non-empty, then it contains a maximal element I. By definition of S, I is not a prime ideal, hence there exist x, x ∈ R such that x, x ∈ I and xx ∈ I.