7

We define a ring homomorphism $F : \mathbb{Q}[x,y]\rightarrow \mathbb{Q}(t)$ by mapping $f(x,y)$ to $f(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2})$. Show $\ker F=(x^2+y^2-1)$.

Proof) Since $(\frac{1-t^2}{1+t^2})^2+(\frac{2t}{1+t^2})^2-1=0$, $(x^2+y^2-1)\subset \ker F$. We note $\ker F$ is a prime ideal since it is the kernel of a ring homomorphism from a commutative ring into a field and $(x^2+y^2-1)$ is also prime as it is irreducible. If $(x^2+y^2-1)\neq \ker F$, since $\dim \mathbb{Q}[x,y] = \dim \mathbb{Q}+2=0+2=2$ (where by dim, we mean the Krull dimensions of rings), $\:\:0\subsetneq (x^2+y^2-1) \subsetneq \ker F$ and $\ker F$ is a maximal ideal. So it now suffices to show that $\ker F$ is not maximal. Suppose it is maximal. Then in $\mathbb{Q}[x,y]/\ker F$, $y + \ker F$ is a unit. Then there exists a polynomial $g(x,y)\in \mathbb{Q}[x,y]$ such that $yg(x,y) + \ker F = 1 + \ker F$. So we have $yg(x,y)-1\in \ker F$. But $\frac{2t}{1+t^2}g(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2})-1$ cannot be $0$ (if so, by clearing the denominators, $2th(t)=(1+t^2)^n$ for $n\in \mathbb{N}$, $h(t)\in \mathbb{Q}[t]$ but the irreducible element $t$ in $\mathbb{Q}[t]$ cannot divide $(1+t^2)^n$), a contradiction.

Could anyone read my proof and let me know any wrong point? If the proof is wrong, I hope you could give me any help. I tried more elementary way to prove the statement and failed.

darij grinberg
  • 17,777
  • When you say "$f(x,y)$ to $f(\dots, \dots)$", what is $f$ and why does it appear on the right of the equal sign? Perhaps it is more clear to say $F(x) = \dots$ and $F(y) = \dots$? – Eric Towers Aug 23 '19 at 01:40
  • By $f(x,y)$, I meant a polynomial in $x,y$ with rational coefficients so $f(x,y)$ is in $\mathbb{Q}[x,y]$. –  Aug 23 '19 at 01:44
  • It looks good to me, but I'd probably add something about $F$ descending to an injective map $Q[x,y]/\ker F\to Q(t)$. From this you can carry on evaluating $yg(x,y)-1$ under $F$. – user347489 Aug 23 '19 at 02:03
  • 1
    @user682705: It looks right. It reminds me of the substitution $t=tan(a/2)$ (TRIGONOMETRIC SUBSTITUTION In INTEGRATION) and then $x:=cos(a)=\frac{1-t^2}{1+t^2}$ and $y:=sin(a)=\frac{2t}{1+t^2}$. ($x^2+y^2-1=0$) – E.R Aug 23 '19 at 03:24
  • 3
    Hint to a simpler proof: We want to show that each $u \in \ker F$ is a multiple of $x^2+y^2-1$. Well, let $u \in \ker F$. We must prove that $x^2+y^2-1 \mid u$. Considering $u$ as a polynomial in $y$ over $F\left[x\right]$, we can divide $u$ with remainder by the monic polynomial $x^2+y^2-1$. We thus obtain $u = q \cdot\left(x^2+y^2-1\right) + r$, where $\deg_y r \leq 1$. Consider these $q$ and $r$. It suffices to show that $r = 0$. Write $r$ in the form $r = a\left(x\right) + b\left(x\right) y$ for two univariate polynomials $a$ and $b$. (We can do this, since $\deg_y b \leq 1$.) Now, ... – darij grinberg Aug 23 '19 at 11:24
  • 3
    ... note that $u - r = q \cdot \left(x^2+y^2-1\right) \in \ker F$ (since $x^2+y^2-1\in \ker F$), and thus $F\left(r\right) = F\left(u\right) = 0$. But applying $F$ to both sides of the equality $r = a\left(x\right) + b\left(x\right) y$, we obtain $F\left(r\right) = a\left(\dfrac{1-t^2}{1+t^2}\right) + b\left(\dfrac{1-t^2}{1+t^2}\right) \cdot \dfrac{2t}{1+t^2}$. Hence, $a\left(\dfrac{1-t^2}{1+t^2}\right) + b\left(\dfrac{1-t^2}{1+t^2}\right) \cdot \dfrac{2t}{1+t^2} = F\left(r\right) = 0$. Now comes the punchline: Substitute $-t$ for $t$ in this equality, and add together. Do you see it now? – darij grinberg Aug 23 '19 at 11:26

0 Answers0