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Zach's question at Yahoo! Answers (Field with 25 elements)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Zach,

$(a)$ The field $\mathbb{Z}_5$ has 5 elements and consider $p(x) = x^2 + 2\in\mathbb{Z}_5[x]$. This polynomial has no zeroes in $\mathbb{Z}_5$ and a quadratic polynomial without zeroes is irreducible. Hence, $F=\mathbb{Z}_5[x]\;/<x^2+2>$ is a field. But every class has one and only one representative of the form $ax+b$ with $a,b\in\mathbb{Z}_5 $. This implies $\#(F)=5\cdot 5=25.$

$(b)$ According to the theory of field extensions, the minimal polynomial of $\sqrt{5}$ is $f(x)=x^2-5$, so a basis of $[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]$ is $B=\{1,\sqrt{5}\}$. As a consequence, $\mathbb{Q}(\sqrt{5})=\{a+b\sqrt{5}:a,b\in\mathbb{Q}\}$.