Field theory MOC

Adjoining a root to a field

Let ๐พ be a field and ๐‘“(๐‘ฅ) โˆˆ๐พ(๐‘ฅ) be a nonzero irreducible polynomial. Then

๐พ(๐›ผ):=๐พ[๐‘ฅ]โŸจ๐‘“(๐‘ฅ)โŸฉ

is a simple extension field of ๐พ, with primitive element ๐›ผ =๐œ‹(๐‘ฅ). Moreover, if ๐ฟ :๐พ is a field extension so that ๐‘“(๐‘ฅ) has a root in ๐ฟ, then we have a tower of field extensions ๐ฟ :๐พ[๐›ผ] :๐พ.1 #m/thm/field

Proof

Since ๐พ[๐‘ฅ] is a Euclidean domain and thus in particular a PID. By Maximal ideal iff prime ideal in a PID, it follows โŸจ๐‘“(๐‘ฅ)โŸฉ is maximal and thus ๐พ(๐›ผ) as defined is indeed a field (๐‘…/๐ผ for commutative ๐‘… is a field iff ๐ผ is maximal). Let ๐œ‹ :๐พ[๐‘ฅ] โ† ๐พ(๐›ผ) denote the projection. Then

๐‘“(๐œ‹(๐‘ฅ))=๐œ‹(๐‘“(๐‘ฅ))=0.

Since all we adjoined was ๐›ผ =๐œ‹(๐‘ฅ), this is indeed simple.

Now suppose ๐ฟ :๐พ is an extension with ๐‘“(๐›ฝ) =0, ๐›ฝ โˆˆ๐ฟ, so the Evaluation map ๐œ–(๐›ฝ) :๐พ[๐‘ก] โ†’๐ฟ vanishes at ๐‘“(๐‘ฅ), whence โŸจ๐‘“(๐‘ฅ)โŸฉ โ‰คkerโก๐œ–(๐›ฝ) and thus by the universal property of quotients there is a unique homomorphism

๐‘—:๐พ(๐›ผ)โ†’๐ฟ

which gives the desired tower of extensions.


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Footnotes

  1. 2009. Algebra: Chapter 0, ยงV.5.2, pp. 283โ€“284 โ†ฉ