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In my algebra class, we proved that a quotient ring $F[x]/(f(x))$ is a vector space over $F$ and $\dim_F F[x]/(f(x)) = \deg f(x).$ I am attempting to use these facts to prove that the field $U = (GF(11)[x])/(x^2+1)$ has $121$ elements.

There is an identity that I found on wikipedia (http://en.wikipedia.org/wiki/Dimension_%28vector_space%29):

$|V| = |F|^{\dim V}$ where $V$ is a vector space over a field $F,$ that makes the proof very simple:

Since $U$ is a vector space over the finite field $\mathbb{Z}/11\mathbb{Z}:$

$|U| = |\mathbb{Z}/11\mathbb{Z}|^{\dim U} = |\mathbb{Z}/11\mathbb{Z}|^{\deg x^2 + 1} = (11-1)^2 = 10^2 = 100 \ne 121.$

Am I using this identity wrongly? We also did not prove this identity in my class, so I do not want to use it without a proof first. How do we prove that $|V| = |F|^{\dim V}$ for a finite dimensional vector space $V$ over $F$?

Note: I am well aware of the duplicate Prove that $\mathbb F_{11}[x]/(x^2+1)$ has 121 elements, I'm just doing the problem a little differently.

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St Vincent
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2 Answers2

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Take a basis in $V$. Then the vector to coordinate mapping is a bijection between $V$ and $F^{\dim V}$. Since $F^{\dim V}$ is a finite set, the space $V$ is a finite set as well, and both have the same number of elements.

user26857
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daw
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I believe the mapping can be constructed as follows. Let $\mathbf e_i$ be the standard basis vector for $F^{\dim V}$, for example, $\mathbf e_2=(0,1_F,0,\ldots ,0)$.

Then, let $\mathcal B=\{v_1,v_2,\ldots ,v_{\dim V}\}$ be a basis of $V$.

Then the map $V\rightarrow F^{\dim V}$ defined by $v_i\mapsto \mathbf e_i$ is a Linear Isomorphism.

I did not look at the rest of your question, but I hope this helps.

Andrew TheDark
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