I would like to find a field $F$ and an epimorphism $\varphi\,\colon \mathbb{Z}_5[X]\to F$ with kernel equal to the ideal generated by the (indecomposable) polynomial $X^2+2$. Is it possible that $F=\mathbb{Z}_5$?
Epimorphism onto a field with kernel isomorphic to $(X^2+2)$
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$\begingroup$
abstract-algebra
polynomials
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0Do you know that $K[x]/f(x)$ is a field if $f$ is irreducible over the field $K$? – 2012-02-23
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0Sure, but I am looking for an explicit epimorphism. – 2012-02-23
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0What do you mean by an "explicit epimorphism"? – 2012-02-23
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0A 'concrete' field $F$ together with an epimorphism $\varphi\colon \mathbb{Z}_5[x]\to F$ with kernel equal to $(X^2+1)$. – 2012-02-23
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4Why isn't $\mathbb Z_5[x]/(x^2+2)$ explicit or concrete? You could also view it as $\mathbb Z_5[\sqrt 3]$ – 2012-02-23
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0What is $\mathbb{Z}_5[\sqrt{3}$? – 2012-02-23
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0Do you know what is $\mathbb Q[\sqrt{3}]$? – 2012-02-23
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0Yes, but what this makes no sense for $\mathbb{Z}_5$. – 2012-02-23
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2Tell us what ${\bf Q}[\sqrt3]$ means to you, and we'll show you that ${\bf Z}_5[\sqrt3]$ makes perfectly good sense. – 2012-02-23
1 Answers
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it's really very simple: add an "extra element" (let's call it $a$) to $\mathbb{Z}_5$ with the following property: $a^2 = 3$. this gives us the following 25 elements:
$\{0,1,2,3,4,a,1+a,2+a,3+a,4+a,2a,1+2a,2+2a,3+2a,4+2a,3a,1+3a,2+3a,3+3a,4+3a,4a,1+4a,2+4a,3+4a,4+4a\}$
the epimorphism you seek is given by $\varphi(f(x)) = f(a)$ (the "evaluation map"). for example, if $f(x) = x^3 + x + 1$, then $\varphi(f(x)) = a^3 + a + 1 = (a^2)a + a + 1 = 3a + a + 1 = 1 + 4a$