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Show that $f: \mathbb{Z}[i]\rightarrow \mathbb{Z}_{17}; f(a+bi)= \widehat{a+4b}$ is a ring homomorphism and show that the kernel of $f$ is $(4-i)\mathbb{Z}[i]$.

I was able to show that $f$ is a ring homomorphism by checking the axioms from the definition but I can't manage to find its kernel.

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3 Answers 3

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Directly:

$$a+bi\in\ker f\iff a+4b=0\pmod{17}\iff a=-4b\pmod{17}$$

$$\iff a=-4b+17k\;,\;\;k\in\Bbb Z\iff a+bi=-4b+17k+bi=$$

$$=(4-i)(-b+4k+ki)\in (4-i)\Bbb Z[i]$$

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  • $\begingroup$ Just to be sure can you explain how you get the last step (i.e. $(4-i)(-b+4k+i)$) $\endgroup$ Sep 20, 2016 at 14:11
  • $\begingroup$ @RaducuMihai To reach that I first divided $\;-4b+17k+bi\;$ by $\;4-i\;$ and got that quotient. Now it is just a matter of checking the product in the complex numbers field:$$(4-i)(-b+4k+ki)=4\cdot(-b+4k) -((-1)\cdot k+(4\cdot k-1\cdot(-b+4k)i=$$$$=(-4b+16k+k)+(4k+b-4k)i=4b+17k+bi$$ $\endgroup$
    – DonAntonio
    Sep 20, 2016 at 16:16
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  • $(4-i)\mathbb{Z}[i] \subseteq \ker f$:
    If $a+bi = (4-i)(x+yi)$, then $a=4x+y$, $b=-x+4y$ and $f(a+bi)=a+4b = 4x+y-4x+16y=17y \equiv 0 \bmod 17$.

  • $(4-i)\mathbb{Z}[i] \supseteq \ker f$:
    You need to prove that if $17$ divides $a+4b$ then $a+bi = (4-i)(x+yi)$ for some $x,y \in \mathbb Z$.

This reduces to solving a linear system: $$ \pmatrix {a \\ b} = \pmatrix{\hphantom{-}4 & 1 \\ -1 & 4 } \pmatrix {x \\ y} $$ The solution is $$ \pmatrix {x \\ y} = \frac{1}{17} \pmatrix{4 & -1 \\ 1 & \hphantom{-}4 } \pmatrix {a \\ b} $$ Now $17$ divides $a+4b$ implies $17$ divides $4a+16b=4a-b+17b$ and so $17$ divides $4a-b$. Therefore, $x,y \in \mathbb Z$.

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Consider the unique ring homomorphism $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}/17\mathbb{Z}$ such that $\varphi(n)=[n]$, for $n\in\mathbb{Z}$, and $\varphi(x)=[4]$. I use $[m]$ for the elements of the quotient ring at the codomain instead of your hat.

Since $\varphi(x^2+1)=[4]^2+[1]=[0]$, we have that $(x^2+1)\subseteq\ker\varphi$, so $\varphi$ induces a ring homomorphism $f\colon\mathbb{Z}[i]\to\mathbb{Z}/17\mathbb{Z}$, where $\mathbb{Z}[i]$ is identified in the usual way with $\mathbb{Z}[x]/(x^2+1)$ (by $i=[x]$ in the quotient ring. Check that this induced homomorphism coincides with your $f$.

Such a homomorphism is obviously surjective, because the codomain has no proper subring. Therefore $\ker\varphi$ is a maximal ideal.

Since $4-x\in\ker\varphi$, $\ker\varphi$ is maximal and $4-x$ is irreducible, $\ker\varphi=(4-x)$. Hence $\ker f=(4-x)/(x^2+1)$ which is the same as $(4-i)$ in the identification of $\mathbb{Z}[x]/(x^2+1)$ with $\mathbb{Z}[i]$

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