Factor ring of Gaussian integers is a field

Asked Apr 12 '22 at 16:30

Active Apr 12 '22 at 16:30

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I want to show that $E=\mathbb{Z}[i]/\langle2-i\rangle$ is a field.

To do this, I note that $R/I$ is a field iff $I$ is a maximal ideal.

Moreover, a maximal ideal is a prime ideal in a commutative ring with unity. We have exactly this in the ring of Gaussian integers $\mathbb{Z}[i]$.

Hence, it suffices to show that $\langle2-i\rangle$ is prime or maximal. I am not sure which is easier to show and in either case how to do so. Any suggestions where to begin?

asked Apr 12 '22 at 16:30

maddiemoo

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There's a useful magnitude function $f: \mathbb{Z}[i] \to \mathbb{N}$, $f(x+yi) = x^2+y^2$, with the property that $f(ab) = f(a)f(b)$. If $2-i = ab$, then $f(2-i) = 5 = f(a)f(b)$... – aschepler Apr 12 '22 at 16:40
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$\Bbb Z[i]$ is a PID, so prime and maximal is the same. We know a lot on maximal and prime ideals then, see for example this duplicate. – Dietrich Burde Apr 12 '22 at 16:47
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Also see again this duplicate which shows that $\Bbb Z[i]/(2-i)\cong \Bbb F_5$ is a field. Hence $(2-i)$ is maximal (and prime). – Dietrich Burde Apr 12 '22 at 16:52
Does this answer your question? Quotient ring of Gaussian integers – Dietrich Burde Apr 12 '22 at 16:53
@DietrichBurde To confirm, would I be correct here in saying $E$ is a field because $\langle2-i\rangle$ is a maximal ideal in $\mathbb{Z}[i]$ as $2-i$ is irreducible in $\mathbb{Z}[i]$? – maddiemoo Apr 12 '22 at 17:47
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Yes, this is correct - see also the answer here. – Dietrich Burde Apr 12 '22 at 18:11

Factor ring of Gaussian integers is a field

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