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I Extension fields

  1. May 5, 2017 #1
    I am a little confused about terminology when it comes to extension fields. In my textbook, E is a field extension of F if F is a subfield of E. This is understandable. However, in proving that all polynomials have a zero in an extension field, ##F[x] / \langle p(x) \rangle##, where ##p(x)## is irreducible, is identified as an extension field of ##F##. But how does that match the definition of extension field given above? ##F## isn't a subfield of ##F[x] / \langle p(x) \rangle## at all, but rather isomorphic to a subfield by the isomorphism ##\mu (a) = a + \langle p(x) \rangle##, right?
     
  2. jcsd
  3. May 5, 2017 #2

    fresh_42

    Staff: Mentor

    Yes.

    Now ##F[x]/\langle p(x) \rangle## is an ##F-##vector space with a basis vector ##1##, so the embedding ##\mu## is a canonical one (maybe even natural) and ##F## can be considered as a subfield. Usually we also write the elements of ##E = F[x]/ \langle p(x) \rangle ## by Latin letters like ##v \in E## instead of ##[v]_{\langle p(x) \rangle }## and elements ##\mu(a)## as ##a## instead of ##a\cdot 1##. It is nothing gained by a rigorous notation here, so it's easier to drop all the extra ##\mu \, , \, [\,.\,]_{\langle p(x) \rangle} \, , \, \cdot 1 , \cdot x , \ldots , \cdot x^{\deg p -1}##.

    It's similar to what we do with complex numbers: ##a + i\cdot b## is far more convenient than ##a\cdot 1 + b \cdot x## or ##\mu(a) + \mu(b)x## or even ##a \cdot [1]_{\langle x^2+1 \rangle} + b \cdot [x]_{\langle x^2+1 \rangle}##. As can be seen here, ##F=\mathbb{R}## would have be to mentioned additionally anyhow. Imagine we would always have to speak of representatives of cosets in ##\mathbb{R}[x]/{\langle x^2+1 \rangle}## instead of real numbers in ##\mathbb{C}##.
     
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