then clearly every non-constant polynomial in F[x] has some root in F; in other words, F is algebraically closed. On the other hand, that the property stated here holds for F if F is algebraically closed follows from the previous property together with the fact that, for any field K, any polynomial in K[x] can be written as a product of irreducible polynomials. Polynomials of prime degree have roots If every polynomial over F of prime degree has a root in F, then every non-constant polynomial has a root in F. It follows that a field is algebraically closed if and only if every polynomial over F of prime degree has a root in F. The field has no proper algebraic extension The field F is algebraically closed if and only if it has no proper algebraic extension. If F has no proper algebraic extension, let p(x) be some irreducible polynomial in F[x]. Then the quotient of F[x] modulo the ideal generated by p(x) is an algebraic extension of F whose degree is equal to the degree of p(x). Since it is not a proper extension, its degree is 1 and therefore the degree of p(x) is 1. On the other hand, if F has some proper algebraic extension K, then the minimal polynomial of an element in K \ F is irreducible and its degree is greater than 1. The field has no proper finite extension The field F is algebraically closed if and only if it has no proper finite extension because if, within the previous proof, the term "algebraic extension" is replaced by the term "finite extension", then the proof is still valid. (Note that finite extensions are necessarily algebraic.) Every endomorphism of Fn has some eigenvector The field F is algebraically closed if and only if, for each natural number n, every linear map from Fn into itself has some eigenvector. An endomorphism of Fn has an eigenvector if and only if its characteristic polynomial has some root. Therefore, when F is algebraically closed, every endomorphism of Fn has some eigenvector. On the other hand, if every endomorphism of Fn has an eigenvector, let p(x) be an element of F[x]. Dividing by its leading coefficient, we get another polynomial q(x) which has roots if and only if p(x) has roots. But if q(x) = xn + an − 1xn − 1+ ⋯ + a0, then q(x)