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In class we are learning about finite fields and polynomials in finite fields my professor wrote that

$Z_{5[x]}/f(x)$ where $f(x)=x^3+x+1$

Here P = 5, n = 3 $|F|=5^3=125$ So F has 125 elements

I'm a bit confused as to how he determined this, could someone provide me with a bit of an explanation?

Temirzhan
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1 Answers1

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Your field $K$ has characteristic $p$. Then $K$ is a finite extension of $\mathbb{Z}_5$ of degree 3. Now choose a basis of $K$ over $\mathbb{Z}_5$, Note that every element of $K$ is a linear combination of such elements with coefficients in $\mathbb{Z}_5$. For each coefficient there are 5 choices (elements of $\mathbb{Z}_5$) and so there are $5^{3}$ linear combinations.

oczalevaj
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