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Consider $K[x^2,x^3] \subset K[x]$, where $x$ is an indeterminate over a (zero characteristic) field $K$.

Clearly, $x$ vanishes the following polynomials $\in K[x^2,x^3][T]$:

$f(T)=x^2T-x^3$, $g(T)=T^2-x^2$, $h(T)=T^3-x^3$, etc.

I wish to find the kernel of the surjective homomorphism $K[x^2,x^3][T] \to K[x]$, defined by $T \mapsto x$; in other words, what are the generators of the kernel? (Obviously, $f(T), g(T), h(T)$ are all in the kernel).

user26857
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user237522
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1 Answers1

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Since $T^2-x^2$ is monic, every element in $K[x^2,x^3][T]$ can be written as $(T^2-x^2)Q(T)+AT+B$, where $A,B\in K[x^2,x^3]$ and $Q(T)\in K[x^2,x^3][T]$. Such an element will be in the kernel if and only if $AT+B$ is in the kernel, which means $Ax+B=0$, or $B=-Ax$.

Suppose it is and write $A=a_0+a_2x^2+a_3x^3+\dots+a_nx^n$; then $-B=a_0x+a_2x^3+\dots+a_nx^{n+1}$ and, since $B\in K[x^2,x^3]$, it must be $a_0=0$.

Thus $A=x^2(cx+f(x))$, for some $f\in K[x^2,x^3]$ and $c\in K$, so $$ AT+B=x^2(cx+f(x))T-x^3(cx+f(x))=c(x^3T-x^4)+f(x)(x^2T-x^3) $$

Thus an element in the kernel belongs to the ideal generated by $T^2-x^2$, $x^2T-x^3$ and $x^3T-x^4$.

The converse is clear.

For the problem of division, let's look at it in a more general setting. Let $R$ be a commutative ring and let $f(T)$ be a monic polynomial in $R[T]$. Then, for every polynomial $g(T)\in R[T]$ there exist $q(T)$ and $r(T)$ such that

  1. $g(T)=f(T)q(T)+r(T)$
  2. $\deg r<\deg f$ (where, as usual, the degree of the zero polynomial is $-\infty$).

The proof is the same as in the field case, by induction on the degree of $g$. If $g=0$ there is nothing to prove, so assume $g\ne0$, $g=a_mT^m+a_{m-1}T^{m-1}+\dots+a_0$.

Write $f(T)=T^n+f_0(T)$, with $\deg f_0<n=\deg f$ (it's not restrictive to assume $n>0$, or the result is obvious). If $\deg g<n$ we take $q=0$ and $r=g$. Otherwise we consider $$ g_0(T)=g(T)-a_mT^{m-n}f(T) $$ which has degree less than $m$. By the induction hypothesis, $$ g_0(T)=f(T)q_0(T)+r(T) $$ with $\deg r<\deg f$. Then $$ g(T)=g_0(T)+a_mT^{m-n}f(T)= f(T)(a_mT^{m-n}+q_0(T))+r(T) $$ and we are done.

It's easy to extend the result to polynomials $f$ having invertible leading coefficients, by considering $u^{-1}f$, where $u$ is the leading coefficient: then $g(T)=u^{-1}f(T)q(T)+r(T)$ and so $u^{-1}q$ is the quotient and the remainder is the same.

Let's see uniqueness. Suppose $$ g(T)=f(T)q_1(T)+r_1(T)=f(T)q_2(T)+r_2(T) $$ with $\deg r_1<\deg f$ and $\deg r_2<\deg f$. Then $$ f(T)(q_1(T)-q_2(T))=r_2(T)-r_1(T) $$ If $q_1-q_2\ne0$, then the degree of $f(q_1-q_2)$ is at least equal to the degree of $f$ (because $f$ is monic), which is a contradiction, because the degree of $r_2-r_2$ is less than $\deg f$.

In your setting, $R=K[x^2,x^3]$ and $f(T)=T^2-x^2$.

egreg
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  • Very nice! thanks. I wonder if one must include a polynomial of degree $1$ (like $x^2T-x^3$ and $x^3T-x^4$) in a set of generators for the kernel, or one can find another set of generators for the kernel, each of degree $\geq 2$? – user237522 Jun 09 '15 at 23:28
  • Can you please explain your first claim that every element of $K[x^2,x^3][T]$ has the form $(T^2-x^2)Q(T)+AT+B$? $K[x^2,x^3][T]$ is a noetherian domain which is not a UFD. What kind of division exists there? I only know that over polynomial rings we have http://math.stackexchange.com/questions/32070/what-is-the-algorithm-for-long-division-of-polynomials-with-multiple-variables. – user237522 Jul 01 '15 at 22:07
  • @user237522 I added the proof. – egreg Jul 01 '15 at 22:22
  • Thank you very much!! A really nice result. – user237522 Jul 01 '15 at 22:49
  • @user237522 It's basically the same as for fields. You can easily extend to polynomials with invertible leading coefficients and, over fields this means all nonzero polynomials. – egreg Jul 01 '15 at 22:53
  • $R$ is only needed to be an integral domain (noetherianity is not necessary)? And what if I do not mind about uniqueness? (only existence of such a form, given $f$ with invertible leading coefficient)? – user237522 Jul 01 '15 at 23:01
  • @user237522 Actually, the hypothesis that $R$ is a domain is only necessary for uniqueness. In the existence proof it's not used, as you can see. Noetherianity is irrelevant. – egreg Jul 01 '15 at 23:02
  • That's what I thought after reading your proof, but wanted to be sure I am not missing something. Thanks! – user237522 Jul 01 '15 at 23:04
  • I think that for uniqueness we can even weaken the assumption that $R$ (hance $R[T]$) is an integral domain; it seems that it is enough to demand that $f$ is (in addition of being monic/ having invertible leading coefficient) not a zero-divisor in $R[T]$. Am I right? – user237522 Jul 01 '15 at 23:15
  • @user237522 Actually it works for any commutative ring. ;-) – egreg Jul 01 '15 at 23:28