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We know $\mathbb{Z}[x]/(x^2+1)\cong \mathbb{Z}[i]$. From this, I guess a similar thing happens, $$\mathbb{Q}[x,y]/(x^2+y^2)\cong \mathbb{Q}[y,yi]$$ by a map $F:\mathbb{Q}[x,y]\rightarrow \mathbb{Q}[y,yi]$ mapping $y\rightarrow y, x\rightarrow yi$. I think I could prove the kernel of $F$ is $(x^2+y^2)$ but I am not sure. Is the argument true?

I got this question from this post, What is the field of fractions of $\mathbb{Q}[x,y]/(x^2+y^2)$?. And if the above mapping is true, we get $\mathrm{Frac}(\mathbb{Q}[x,y]/(x^2+y^2))=\mathrm{Frac}(\mathbb{Q}[y,yi])=\mathbb{Q}(y,yi)=\mathbb{Q}(y,i)$

user26857
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  • $T^2+1$ is the minimal polynomial of $i$ and $T^2+y^2$ is the minimal polynomial of $yi$ over $\Bbb{Q}(y)$ and $\Bbb{Q}[y]$. Why aren't you sure of the kernel of your map ? The idea of minimal polynomial is that it is a polynomial with minimal degree with the element as a root, if there is another one which is not a multiple then the element is also a root of the $\gcd$ of both polynomials having a smaller degree, a contradiction. – reuns Aug 23 '19 at 05:16
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    How is the minimal polynomial of an element over a ring defined? I know for a field $F$, if $p(x)$ is a monic irreducible polynomial with $\alpha$, $F(\alpha)\cong F[x]/(p(x))$. Does this also hold for a ring $R$, $R[\alpha]\cong R[x]/(p(x))$? –  Aug 23 '19 at 05:23
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    For an integral domain in the same way as over the fraction field, except the leading coefficient does change something. If the $R$-minimal polynomial is monic then it is the same over $R$ and $Frac(R)$. – reuns Aug 23 '19 at 05:28
  • @reuns Frac(R) is supposed to be a field. Then what means Frac(R)/I? – user26857 Aug 23 '19 at 08:55
  • I meant $R[x]/(I \cap R[x])$ is a subring of $Frac(R)[x]/I$. If $I = pFrac(R)[x]$ and $p$ is monic $\in R[x]$ then $I \cap R[x] = p R[x]$ – reuns Aug 23 '19 at 08:57
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    If $f\in\ker F$, then $f(yi,y)=0$. In order to prove that $f\in(x^2+y^2)$ write $f(x,y)=(x^2+y^2)g(x,y)+a(y)+b(y)x$ by using the long division with respect to $x$. From $a(y)+iyb(y)=0$ you can deduce that $a(y)=b(y)=0$. (Recall that $a(y),b(y)\in\mathbb Q[y]$.) – user26857 Aug 23 '19 at 15:53
  • Thank you very much for the clarification! –  Aug 23 '19 at 18:36

1 Answers1

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The standard monomials of ${\Bbb Q}[x,y]/\langle x^2+y^2\rangle$ are $1,y,y^2,y^3,\ldots$ and $x$ with $x^2=-y^2$ and so $x=\pm iy$. This gives certainly an isomorphism of vector spaces with ${\Bbb Q}[y,iy]$. That it is also an isomorphism of rings requires a proof: $y\mapsto y$ and $x\mapsto iy$.

More insight into standard monomials is provided by Cox, Little, O'Shea ''Using Algebraic Geometry'' and the more elementary text of Cox, Little, O'Shea ''Ideals, Varieties, and Algorithms''.

Wuestenfux
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  • Don't forget $xy^i$ for all $i \geq 0$ (not just $i = 0$). – darij grinberg Aug 25 '19 at 07:25
  • Anyway, using standard monomials is slightly overkill here :) Just note that $x^2+y^2$ is a monic polynomial of degree $2$ in the variable $x$ over $\mathbb{Q}\left[y\right]$. Thus, $\left(\mathbb{Q}\left[y\right]\right)\left[x\right] / \left<x^2+y^2\right>$ is a free $\mathbb{Q}\left[y\right]$-module with basis $\left(\overline{1}, \overline{x}\right)$. Now it is easy to construct your isomorphism directly as a $\mathbb{Q}\left[y\right]$-algebra isomorphism (no need to fall back down to the level of $\mathbb{Q}$-algebras). – darij grinberg Aug 25 '19 at 07:27