I'm just spamming blog posts now.

Let be a commutative ring. is a field iff it has only two ideals, itself and .

Proof. If is a field then any nonzero ideal contains a nonzero element , which generates , which generates the whole field, so . So and are indeed the only two ideals.

If is not a field then there's a noninvertible element and the ideal generated by is neither nor .

— end proof —

(Theorem 2.16 in Jacobson) Let be a field and let be algebraic over with minimal polynomial . Then is:

• a field if is irreducible;
• not even an integral domain if is reducible.

Proof. If is irreducible, we claim has no ideals other than itself and . So suppose we have an ideal of . Then the preimage of under the obvious homomorphism from the polynomial ring to would be an ideal of . Since is a principal ideal domain, as seen in the last post, is generated by an element . Also contains , so is a multiple of . Since is irreducible we must have or (up to a constant factor), which respectively means or , as desired. Thus is a field by the statement earlier in this post.

If is reducible as then and are nonzero in but is zero, so is not an integral domain.

Corollary. If is a field and is embeddable in an integral domain, then is a field.

Proof. As above, is either a field or contains zero divisors, and the latter case is impossible if is embeddable in an integral domain.