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

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A field of order 25 can be constructed using the polynomial p(x) = x^2 + 2 over the field Z_5, which is irreducible and leads to the field F = Z_5[x]/<x^2 + 2>. This results in 25 distinct elements represented as ax + b, where a and b are in Z_5. For the field Q(√5), the minimal polynomial is f(x) = x^2 - 5, indicating that a basis is {1, √5}. Therefore, the elements of Q(√5) can be expressed as a + b√5, where a and b are rational numbers.
Fernando Revilla
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Here is the question:

a) Construct a field of order 25

b) Describe the elements of Q(sqrt5)

Here is a link to the question:

Abstract Linear Algebra? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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, $\mathbb{Q}(\sqrt{5})=\{a+b\sqrt{5}:a,b\in\mathbb{Q}\}$.
 
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