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enter image description here I have a quick question about the allowed function use to show isomorphism between quotient rings. The question assumes the following exercise:

Let $R=\{a+bi|a,b\in \mathbb{Z}\}$, and $J=\{a+bi| 5|a, 5|b\}.$ Show that $J$ is not a maximal ideal in $R$ (Hint: Consider the principal ideal $K=\{a+bi\}$ in $R$)

The question asks: If $R$ and $J$ are as in Exercise 1, show that $R/J \cong \mathbb{Z_{5}}$

I know that we can define $\psi: R\rightarrow \mathbb{Z_{5}}$ by $\psi(a+bi)=[a-2b]_{5}$, but can I use the function $\psi(a+bi)=[a+2b]_{5}$ instead?

Note: it is question 24 in the attached picture which assumes notations from exercises 21(a) and 22.

Thank you in advance.

Seth
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  • Well, you cannot show both statements “$J$ is not a maximal ideal” and “$R/J\cong\mathbb{Z}_5$”, because they contradict each other. – egreg Dec 22 '21 at 10:18
  • @egreg I fixed the question. I think the way I typed it might caused a bit of a misunderstanding. It question assumes the first statement. I am not sure if $[a+2b]{5}$ would make the solution not work instead of $[a-2b]{5}$. – Seth Dec 22 '21 at 10:24
  • Again, if $J$ is as in the stated exercise, then $R/J$ cannot be isomorphic to a field, but $\mathbb{Z}_5$ is a field. – egreg Dec 22 '21 at 10:35
  • @egreg I have fixed the question with attached screenshot of where the exercises came from and also fixed the typing of the post. My profuse apologies for the confusions. – Seth Dec 22 '21 at 10:47
  • We have many prior questions showing how to prove such isomorphisms, e.g. see the linked dupes. – Bill Dubuque Dec 22 '21 at 13:57

1 Answers1

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The text doesn't match the screenshot. You're required to prove that if $K=(2+i)$, then $R/K$ is isomorphic to $\mathbb{Z}_5$.

One way is to consider the map $\psi\colon R\to\mathbb{Z}_5$ defined by $\psi(a+bi)=[a-2b]_5$.

Is this a ring homomorphism? One could check this via the definition, but there's a better way. We can consider $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}_5$ defined by $\varphi(x)=[-2]_5$.

Now we can see that the kernel of this homomorphism contains the ideal $I$ generated by $x^2+1$, because $\varphi(x^2+1)=[-2]_5^2+[1]_5=[0]_5$. Hence this homomorphism induces a ring homomorphism $\psi\mathbb{Z}/I\to\mathbb{Z}_5$ and $\psi(x+I)=[-2]_5$. Now we can “canonically” identify $R$ with $\mathbb{Z}[x]/I$ and we're done, because $$ \psi(a+bi)=\psi(a)+\psi(b)\psi(i)=[a]_5+[b]_5[-2]_5=[a-2b]_5 $$

Can you do the same by mapping $x$ to $[2]_5$? Yes, why not? Repeat the argument and you're done.

Or consider that conjugation is an automorphism of $R$ and if $\psi'(a+bi)=[a+2b]_5$, then $$ \psi'(a+bi)=\psi(\overline{a+bi}) $$ so $\psi'=\psi\circ j$, where $j$ is conjugation.

egreg
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  • sorry I checked and fixed the typing to align with the screenshot. THe solution from the text reads as follows:

    "As in Exercise 22 define $\psi:R\rightarrow \mathbb{Z_{5}}$ by $\psi(a+bi)=[a-bi]{5}$. The work done there already implies that $\psi$ is a surjective homomorphism. If $a+bi$ is in the kernel then $[a-bi]{5}=[0]_{5}$, so that $a=2b+bq$ for some $q\in \mathbb{Z}$. Therefore $a+bi=(2+i)b+5q=(2+i)\cdot(b+(2-i)q)$ lies in the ideal $(2+i)$. Consequently the kernel is $(2+i)$ and the first isomorphism theorem can be applied."

    What I like to know is.....

     – Seth Dec 22 '21 at 11:07
  • #egreg ...do I really need a conjugation map. I often see conjugation map used when it involves complex numbers or proving things involving binomial theorem identities. – Seth Dec 22 '21 at 11:08
  • @SethMai It's a basic tool for complex numbers: why wouldn't you use it? – egreg Dec 22 '21 at 11:09
  • I know complex analysis, but when i see it used, there seems to be no rhyme or reasons for its use. All i found in the textbook literature is that conjugation map has to do with complexification of a real vector spaces like in Berger's book:

    https://www.google.ca/books/edition/Geometry_I/5W6cnfQegYcC?hl=en&gbpv=1&dq=complexification+of+vector+spaces,+marcel+berger&pg=PA145&printsec=frontcover

     – Seth Dec 22 '21 at 11:13
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    @SethMai Conjugation is an automorphism of $\mathbb{C}$ and it induces an automorphism of $\mathbb{Z}[i]$. What's the problem? Don't look too far: this is not “complex analysis”, but basic algebra. – egreg Dec 22 '21 at 11:23
  • ah okay okay. Thank you for the explanation. How can I miss this little detail. Sometimes, when I do math, all the little things just don't come to the forefront readily. – Seth Dec 22 '21 at 11:31