Possible Cardinality of a Field
Solution 1
In priciple, yes. $F$ is a vector space over $\mathbb F_p$ and hence in the finite case it is in bijection with some $\mathbb F_p^n.$ Of course $|\mathbb F_p^n|=|\mathbb F_p|^n=p^n$.
Solution 2
I did not manage to follow your argument however there is a very simple argument here:
Since you already noted that the prime field of $F$ is $\mathbb{F}_{p}$ (up to isomorphism) all you have to recall is that $F$ is a vector space over its prime field hence $$|F|=|\mathbb{F}_{p}|^{dim_{\mathbb{F_p}}(F)}=p^{dim_{\mathbb{F_p}}(F)}$$
AsinglePANCAKE
Updated on September 29, 2020Comments
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AsinglePANCAKE about 3 years
The following question struck me as pretty interesting: Let $\Bbb F$ be a field of characteristic $p$ (a prime, of course). I'm then asked to show that $|\mathbb{F}| = p^n$ for some $n\geq 1$.
Here's my intuition. Certainly we know that the prime subfield of $\mathbb{F}$ has order $p$. Now if there's an element (treating $\mathbb{F}$ now as a vector space over itself) independent from it, we have the $Span\{1,a_1\}$ as the usual set of linear combinations of $1$ and $a_1$. And any element of a field of characteristic $p$ added to itself $p$ times is $0$, so now we have $p^2$ possible linear combinations. And so on, arguing inductively. Is this argument kosher? Or does more need to be said to make it rigorous?
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Gerry Myerson about 11 yearsThe question is answered math.stackexchange.com/questions/53877/… and elsewhere on this site.
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AsinglePANCAKE about 11 years@GerryMyerson Might you point me to the elsewhere? That answer uses language a bit over my head unfortunately...Apologies for the reposting...
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Gerry Myerson about 11 yearsmath.stackexchange.com/questions/183462/…? which you can find by looking at the list headed Linked on the right side of the page in my earlier comment. Look around, check out some of the links, you won't break anything.
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AsinglePANCAKE about 11 years@GerryMyerson Except perhaps the patience of veteran stack-exchangers! Many thanks!
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lhf about 11 yearsYou mean, a finite field.
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AsinglePANCAKE about 11 yearsIf you don't mind, might you elaborate a bit on the string of equalities? They don't resonate with my n00b-self as obvious...
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Belgi about 11 yearsDo you know that up to isomorphism there is only one vector space of dimesnion $n$ over a given field ?
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AsinglePANCAKE about 11 yearsUmm, this makes sense, but I don't think I've seen it before.
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AsinglePANCAKE about 11 yearsAh yes, that's the answer I was poorly articulating above. Thank ya!