## Algebraic field extensions exist and are unique

We will now prove some basic facts about algebraic field extensions.

Let be a field and let be an irreducible monic polynomial of positive degree over . Define . We claim the following.

- is an extension of .
- contains an element which is algebraic over and whose minimal polynomial is .
- If is an extension of which contains an element which is algebraic over with minimal polynomial , then there is an injective ring homomorphism .

Note that is a Euclidean domain, hence a principal ideal domain. Since is irreducible, it is also prime, so that is a prime ideal. Since is a principal ideal domain, is maximal. So is a field. Now let be the map . If , then . Since every nonzero element of has degree at least 1, , so that . Hence is (isomorphically) an extension of .

Let denote the image of ; . Certainly .

Now suppose exists. By the universal property of polynomial rings, we can define a homomorphism by fixing and mapping . Note that , so that . Conversely, if , then , so that the minimal polynomial divides . Thus . By the first isomorphism theorem for rings, given by is a well defined ring homomorphism. Suppose ; then , so that . So divides , and we have . So is injective.

This allows us to characterize as the “smallest” extension of which contains a root of . Moreover, in this previous exercise we showed that the elements of are uniquely of the form for some of degree less than the degree of .

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