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Math Help - Quotient ring, field, irreductible

  1. #1
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    Quotient ring, field, irreductible

    Let K be a field and f \in K[x]. Prove that K[x]/ \langle f \rangle is a field if and only if f is irreductible.

    How can I build a field with nine elements?

    Prove that R[x]/ \langle x^2 + 1 \rangle \cong \mathbf{C}

    Thanks!
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  2. #2
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    Quote Originally Posted by roporte View Post
    Let K be a field and f \in K[x]. Prove that K[x]/ \langle f \rangle is a field if and only if f is irreductible.
    Say that f is irreducible then we need to show for any [g] \in K[x]/(f) with [g]\not = [0]= (f) (i.e. non-zero) we can find [h] such that [g][h] = [gh]= [1]. Now since f,g are relatively prime since f is irreducible it means there exists q,h so that fq+hg=1 and so hg\equiv 1 ~ (f) i.e. [hg] = [1].

    Now you try proving the converse by assuming that f is reducible.

    How can I build a field with nine elements?
    \mathbb{F}_3 / (x^2+1)

    Prove that R[x]/ \langle x^2 + 1 \rangle \cong \mathbf{C}
    Define \theta : R[x] / (x^2+1) \to \mathbb{C} by \theta(1) = 1 and \theta (x) = i. Now show this extends to a function which is in fact a field isomorphism.
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