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Is the map $f: \mathbb C \to \mathbb Z$, $f(a+bi)=a$, for $a,b$ in $\mathbb Z$, a homomorphism of rings?

Any one could give me some ideas please?

  • 3
    It seems like the domain of this function is really the gaussian integers $\mathbb{Z}[i]$, and not all of $\mathbb{C}$2012-03-25

3 Answers 3

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No.

If it was a homomorphism, then $f[(a+bi)(c+di)]=f(a+bi)f(c+di)=ac$

But $f[(a+bi)(c+di)]=f[(ac-bd) + (ad+bc)i]=ac-bd$

  • 0
    thank you for your answer, really helpful i understand the homomorphism of rings now.2012-03-31
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Verify the properties of a ring homomorphism $f$:

$f((a + ib)(c + id)) = f(ac + i(ad + bc) - bd) = ac - bd \neq f(a + ib)f(c + id) = ac$.

So the answer is no.

2

With even less computation: Since $i^2=-1$ the set $\{a+bi | a=0\}$ is not an ideal in $\mathbb Z[i]$