fields

**ulysses123** · Sep 3rd 2010, 07:41 PM

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?

**NonCommAlg** · Sep 4th 2010, 01:42 AM

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.

**ulysses123** · Sep 4th 2010, 02:26 AM

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.

**NonCommAlg** · Sep 4th 2010, 03:34 AM

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 $|V|=|k|^{\deg f}.$

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