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I'm trying to solve the following problem:

Prove that $\mathbb{Q}[x]/(x^2+1)$ and $\mathbb{Q}[y]/(y^2+2y+2)$ are isomorphic.

First of all, I tried to construct homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[y]/(y^2+2y+2)$ with $\mathbb(x^2+1)$ as a kernel, and also the other way around, but without much success.

Then, I've seen the following comment (specifically the part where he proves that $\mathbb{Q}[x]/(x^2-1)$ is isomorphic to $\mathbb{Q}[x]/(x^2-2x)$) : https://math.stackexchange.com/a/57766/808138, and since $y^2+2y+2 = (y+1)^2+1$, i realized that the problems are similar, but there is a part that confuses me: "Since $x^2-2x = (x-1)^2-1$, the equivalence class of $x$ in $\mathbb{Q}[x]/(x^2-1)$ should behave like the equivalence class of $x−1$ in $\mathbb{Q}[x]/(x^2-2x)$". I see that if I define $f(a) = a^2-1$ then $f(a-1) = (a-1)^2-1 = a^2-2a$, but I don't really get why mentioned equivalence classes behave same in respected quotient rings.

EDIT: I rewrote $\mathbb{Q}[x]/(x^2+2x+2)$ as $\mathbb{Q}[y]/(y^2+2y+2)$, so things look more clear. Specifically, when I tried to construct the homomorphism mentioed in first paragraph, I was planning on using The First Isomorphism Theorem, but really didn't have any idea how to construct a useful homomorphism from polynomial ring to polynomial quotient ring. Maybe I am just inexperienced with these type of problems, so any more hints would be greatly appreciated.

user26857
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Ranko
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    Please, for goodness sake, rewrite the second ring as $\mathbb Q[y]/(y^2+2y+2)$! It's way too confusing trying to figure out which ring you're in when they both use $x$. – JonathanZ May 27 '23 at 01:35
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    And once you've done that, you can see that in the first one you mod out by $(x)^2+1$, and in the second one you mod out by $(y+1)^2+1$. This should give you a big hint as to how $x$ and $y$ should be related, and tell you exactly what the homomorphism you were looking for in your first paragraph should look like. Maybe add more details there? What did you try, and what problems did you hit? – JonathanZ May 27 '23 at 01:44
  • Sorry for confusion, I thought things looked clear. Hopefully this edit resolves any issues. I still don't get it how equality you have mentioned should give me a big hint how x and y are related. – Ranko May 27 '23 at 12:25
  • Thanks for the rewrite. I'm sure the problem was presented to you with $x$'s in both places. – JonathanZ May 27 '23 at 12:27
  • Now, when you have an isomorphism, say $f: A\to B$, then $a \in A$ should have the same algebraic properties as $f(b) \ in B$. In your first ring, $x$ had the property that if you square it and add 1, you get zero. So $x$ should map to an element that has the same property in $B$. Can you name one? – JonathanZ May 27 '23 at 12:33
  • Yes, it should an element $y+1$. I was just confused with what author meant by the words that are cited in my post. Why should those equivalence classes behave in a mentioned way? – Ranko May 27 '23 at 12:51
  • When you take a quotient space, the elements are equivalence classes. "The equivalence class of $x$ in $\mathbb{Q}[x]/(x^2-1)$" is what $x$ maps to under the quotient, i e. $x+(x^2-1)$. – JonathanZ May 27 '23 at 12:56
  • So, if $f$ is an isomorphism, then $f(x+(x^2-1)) = y-1+(y^2-2y)$, i.e. equivalnce class of $x$ in $\mathbb{Q}[x]/(x^2-1)$ is same as equivalnce class of $y-1$ in $\mathbb{Q}[y]/(y^2-2y)$? – Ranko May 27 '23 at 13:20
  • That is correct, except you should say "maps to (under $f$)" instead of "is same as". – JonathanZ May 27 '23 at 13:24
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    I guess these things are obvious, but I am new to abstract algebra and I am self-studying it, so I really appreciate nice explanations like this. Thank you so much! – Ranko May 27 '23 at 13:27
  • No problem. There's a lot of new terminology, and even after you learn to read about it, it's a different skill learning how to use the terms yourself. Good luck! – JonathanZ May 27 '23 at 13:30
  • Actually, I have one more question. You said that if $f: A\to B$ is isomorphism, then $a \in A$ should have the same algebraic properties as $f(b)$ in $B$. Does it have to be $b = a$? – Ranko May 27 '23 at 15:05
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    Oh, good catch. I should have typed "... as $f(a)$ in $B$". And looking back at it, "$f(a)$ should have the same properties as $a$" is closer to the way I think about it. $f$ is a bijection, so really one can think of it going either way, but in our case we had an element in $A$ and were looking for something to map it to. – JonathanZ May 27 '23 at 15:14

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The ideals $(x^2+1)$ and $(x^2+2x+2)$ are maximal in $\Bbb Q[x]$, so the quotients are fields. If you call $i$ a root of $x^2+1$, $-1+i$ is a root of $x^2+2x+2$. Both fields are $\Bbb Q[i]=\Bbb Q(i)$.

ajotatxe
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  • Why is it important that the quotients are fields? If I understand you correctly, you are telling me that roots of both polynomials can be expressed involving $i$, so they are both isomorphic to $\mathbb{Q}(i)$? – Ranko May 27 '23 at 12:36
  • @Ranko that is correct. As theyre both maximal they both mod out to a field and the same field as $i \not\in\Bbb{Q}$. But $1 \in \Bbb{Q}$ so theyre both isomorphic to $\Bbb{Q}(i)$. – homosapien May 27 '23 at 17:29