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I wanted to show that $k[X,Y]/(X^2-Y) \cong k[T]$.

However, i do explicitely not want to define an inverse map but use the first isomorphism theorem (factoring through the kernel).

Now, clearly \begin{align}\Phi\colon k[X,Y]&\to k[T] \\ X&\mapsto T \\ Y &\mapsto T^2\end{align}

is surjective and $(X^2-Y)\subseteq \ker \Phi$. I have trouble showing that $\ker \Phi \subseteq (X^2-Y)$.

My attempt:

Let $F(X,Y)\in \ker \Phi$. By polynomial division with remainder, we know there exist unique polynomials $Q(X,Y)$ and $R(X,Y)$ s.t. $$F(X,Y) = (X^2-Y)Q(X,Y)+ R(X,Y),\quad \operatorname{deg}_X R(X,Y) < 2$$

Thus $R$ is a polynomial of the form $$R(X,Y) = A(X,Y)+B(X,Y)X$$ with $A,B\in k[X,Y]$.

Since $\operatorname{deg}_X R(X,Y) < 2$, both $A$ and $B$ do not depend on $X$, i.e.

$$R(X,Y) = A(0,Y) + B(0,Y)X$$

By assumption, $F\in \ker\Phi$, hence $F(T,T^2) = 0$, therefore $R(T,T^2) = 0$, which yields

$$0 = A(0,T^2) + B(0,T^2)T$$

What's left showing is that $$0 = A(0,T^2) + B(0,T^2)T$$ implies that $A = 0$ and $B = 0$, but I'm not sure, how to proceed.

My naive guess:

If $B\not=0$, we would possibly obtain some polynomial in $T$ of degree $\ge 3$ but since $R$ was supposed to be of degree $\operatorname \deg_X R < 2$, we must conclude that $B(0,T^2) = 0$.

Which leaves us with $$0 = A(0,T^2)$$ and then we're done.

My question:

Is this correct? If not, please help me showing my errors. I really want to understand this thoroughly.

Thank you for your help!

Henno Brandsma
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Zest
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  • Did you already compare with this post? It has an answer by the OP himself, which could clarify some of your doubts. Perhaps you can also answer your question then yourself. – Dietrich Burde Oct 27 '21 at 10:10
  • Thanks for the comment @DietrichBurde. I feel like the proposed solution in that post is slightly different to what i've tried. I would really like to try solving it in the way I've tried above since we had similar problems in the past. I feel like I'm almost there, i just need the final argument. – Zest Oct 27 '21 at 10:18
  • Sure, it is always good to follow up the own way, but it is also always good to look up the standard solution for comparison. This is often useful for knowing the "usual" techniques for such exercises. See for example this post and many others here. So, the "usual" technique is the first isomorphism theorem. – Dietrich Burde Oct 27 '21 at 10:21
  • Nono, it's not an exercise/homework. We solved this problem together in an exercise session last week but we did it by defining an inverse map. I just wanted to prove this (for my own) with the proposed different approach. – Zest Oct 27 '21 at 10:33
  • Yes, sure. I just wanted to say, I finally understood you. All you need to show (and to ask) is that $\ker(\phi)=(y-x^2)$. And this you can find also on this site. – Dietrich Burde Oct 27 '21 at 10:34
  • Yes, that's what i want to prove. I just would like to fully understand what's left in my proposed attempt. We had similar exercises in the past where we took this route but I had issues following them. This is the reason i wanted to really solve this problem in this way. It's less about solving it, it's more about understanding this particular approach. – Zest Oct 27 '21 at 10:36

2 Answers2

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$f(T^2)+g(T^2)T=0$ implies $f=g=0$ for the following reason: we add a polynomial containing only even powers of $T$ with a polynomial containing only odd powers of $T$, so the coefficients of $f$ and $g$ are all zero.

user26857
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  • Thanks a lot for your answer, would you mind elaborating why the different powers imply that the coefficients are zero? – Zest Oct 27 '21 at 13:41
  • Is it because if they had mixed powers, we could group them accordingly? – Zest Oct 27 '21 at 13:56
  • Just recall that a polynomial is zero iff all its coefficients are zero. In our case the coefficients of $f$ are the coefficients of even powers of $T$ while the coefficients of $g$ are the coefficients of odd powers of $T$. – user26857 Oct 27 '21 at 17:53
  • Take an example: $f(t)=a+bt$ and $g(t)=c+dt$. Write $f(t^2)+g(t^2)t=0$ and see what's going on. – user26857 Oct 27 '21 at 17:59
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    Ahh, okay now i see. Thanks a lot. I was somehow ignoring that $R(T,T^2) = 0$ means that $R$ is indeed the zero polynomial and not just for some (particular) $T$. Thanks for your patience and help. – Zest Oct 27 '21 at 18:11
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The answer by @user26857 answers your direct question.

I think the argument is slightly easier if you consider $F(X,Y)$ as a polynomial in $Y$ over $k[X]$ and write $$F(X,Y) = (X^2 - Y) Q(X,Y) + R(X,Y) \quad\text{with}\quad \deg_Y R(X,Y) < 1.$$ Then, in fact, $R(X,Y)$ is just a polynomial $R(X)$ in $X$.

Continuing with the same argument, substituting $T$ for $X$ and $T^2$ for $Y$ should give $0$ because $F$ is in the kernel, so $R(T) = 0$. Therefore also $R(X) = 0$.

Magdiragdag
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  • Ah, that's really helpfull, thanks a lot. I was wondering whether we would actually do the polynomial division for $Y$ as you proposed. Thanks @Magdiragdag. – Zest Oct 27 '21 at 13:42
  • Would you mind sharing why the argument in @user26857 regarding the powers implies that we have only zero coefficients? – Zest Oct 27 '21 at 16:22
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    That argument is in user26859's answer already: in the polynomial $f(T^2) + g(T^2)T$ all the monomials with even powers of $T$ are in $f(T^2)$ and all the monomials with odd powers of $T$ are in $g(T^2)T$. Because it is the zero-polynomial, all coefficients are zero. In particular, the coefficients of all even powers of $T$ are zero (so $f(T^2) = 0$) and the coefficients of all odd powers of $T$ are zero (so $g(T^2)T = 0$). – Magdiragdag Oct 27 '21 at 16:39