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Math Help - fields

  1. #1
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    fields

    the set of real numbers is an infinite field.
    so R[x] is infinite
    does this mean that R[x]/<p>
    for some irreducible polynomial p will always be an infinite field or are there exceptions?
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  2. #2
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    the answer is yes. in fact \mathbb{R}[x]/ \langle p \rangle is either \mathbb{R} or \mathbb{C}. the reason is that the degree of an irreducible polynomial over \mathbb{R} is at most 2.

    also note that for any field k and any non-constant polynomial f \in k[x], the ring k[x]/\langle f \rangle contains a copy of k. the reason is that the map \varphi : k \longrightarrow k[x]/\langle f \rangle defined by

    \varphi(\alpha)=\alpha + \langle f \rangle is an injective ring homomorphism. so if k is infinite, k[x]/\langle f \rangle will be infinite too.
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  3. #3
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    What about if K[X]=Q[X]
    this is always stated as "countable" because there is a bijective map onto the natural numbers, but there are still atually infinitely many elements, so would this also be infinite? to my understanding the only finite fields of this nature are the galois fields and all other finite fields will be isomomorphic to these anyway.
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  4. #4
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    i alreday answered your question! for any field k (so you may choose k=\mathbb{Q}) and any non-constant polynomial f \in k[x], not necessarily irreducible, the ring V=k[x]/\langle f \rangle contains

    a copy of k. so you may consider V as a vector space over k. it is clear that 1 \leq \dim_k V \leq \deg f. of course if f is irreducible over k, then  \dim_k V = \deg f.

    as a result, if k is infinite, then |V|=\text{card}(k). thus if k is a countable (resp., uncountable) field, then V is a countable (resp., uncountable) ring.

    if k is finite, then |k| \leq |V| \leq |k|^{\deg f}. if k is finite and f is irreducible, then |V|=|k|^{\deg f}.
    Last edited by NonCommAlg; September 4th 2010 at 04:46 AM.
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