Ytukyg

Let $f=f(t),g=g(t) in mathbb{C}[t]$ be two separable polynomials of degrees $deg(f)=n geq 2$ and $deg(g)=m geq 2$, namely, $f$ has $n$ distinct roots and $g$ has $m$ distinct roots.

Denote $d=d(t)=gcd(f(t),g(t))$, and assume that $deg(d)=l geq 1$.

Assume that $d(t)=(t-gamma_1)cdots(t-gamma_l)$,
so we can write
$f(t)=(t-gamma_1)cdots(t-gamma_l)(t-alpha_1)cdots(t-alpha_{n-l})$
and
$g(t)=(t-gamma_1)cdots(t-gamma_l)(t-beta_1)cdots(t-beta_{m-l})$,
where
${alpha_1,ldots,alpha_{n-l},beta_1,ldots,beta_{m-l},gamma_1,cdots,gamma_l}$ are distinct.

Further assume that $mathbb{C}(f,g)=mathbb{C}(t)$.

Are there nice (necessary and sufficient) conditions for $mathbb{C}[f,g]=mathbb[t]$?

Examples:

(1) $f=t(t-1)(t+1)=t(t^2-1)=t^3-t, g=t(t-i)(t+i)=t(t^2+1)=t^3+t$
We have $f-g=t^3-t-(t^3+t)=-2t$, so $mathbb{C}[f,g]=mathbb{C}[t]$.

(2) $f=(t+1)t(t-1), g=(t+1)(t+2)(t-2)$. By this we get that $mathbb{C}(f,g)=mathbb{C}(t)$ (since $deg(d)=1$).
Here $mathbb{C}[f,g]neq mathbb{C}[t]$, since the D-resultant of $f$ and $g$ is $s(s+1)$, and a necessary and sufficient condition for $mathbb{C}[f,g]=mathbb{C}[t]$ is that the D-resultant of $f$ and $g$ is a non-zero scalar (namely, $in mathbb{C}^{times}$), see Theorem 2.1.

Several ideas:

• One plausible answer is to adjust that Theorem 2.1 to our special case; however, computing the D-resultant is very complicated for higher degrees of $f$ and $g$, or maybe I am missing something and the D-resultant is nicer that I thought?




• What about sub-resultants? (I think I can obtain a nice condition involving sub-resultants) but again in higher degrees the computations are complicated.




• What about SAGBI bases; it seems that ${f,g}$ is not a SAGBI basis, but does this tell something intersting?




• By Abhyankar-Moh-Suzuki theroem, $m$ must divide $n$ or vice versa.




• A related question is this question, which asks: For which $f,g in k[t]$,
  $k[f,g]$ is integrally closed in its field of fractions $k(f,g)=k(t)$?
  Notice that, in our case, if $mathbb{C}[f,g]$ is integrally closed in $mathbb{C}(f,g)$, then, since here $mathbb{C}(f,g)=mathbb{C}(t)$, we obtain that $t$, which is obviously integral over $mathbb{C}[f,g]$, already belongs to $mathbb{C}[f,g]$, hence $mathbb{C}[f,g]=mathbb{C}[t]$.




• See also this question.

Special cases are also wellcome.

Plausible special cases:

(1) $deg(d)=1$: Notice that in each of the two examples $deg(d)=1$, but this does not help in determinig if $mathbb{C}[f,g]=mathbb{C}[t]$ or not, because in the first example we have $mathbb{C}[f,g]=mathbb{C}[t]$, while in the second example we have $mathbb{C}[f,g]neqmathbb{C}[t]$.
Perhaps I am missing a nice criterion that distinguishes between those two examples?

(2) Special forms of $f$ and/or $g$, for example $f(t)=t^n+at^{tilde{n}}+b$ and $g(t)=t^m+ct^{tilde{m}}+d$.

Any hints and comments are welcome!

As promised in the comments, I now prove the following claim:

Claim: Let $k$ be an algebraically closed field of characteristic zero and let $f=f(t),g=g(t) in k[t]$ be such that $f'$ and $g'$ are not simultaneously zero (this, in some cases, implies that $k(f,g)=k(t)$).
Then the following two conditions are equivalent:

(i) $k[f,g]=k[t]$.

(ii) For all $lambda,mu in k$, we have $deg(gcd(f-lambda,g-mu)) leq 1$.

Proof:

Lemma (it is necessary to assume that $k$ is algebraically closed): $k[f,g]=k[t]$ if and only if $f'$ and $g'$ are not simultaneously zero and
$psi: t mapsto (f(t),g(t))$ is injective.

(i) implies (ii):

Denote $F_{lambda}=F_{lambda}(t):=f(t)-lambda$ and $G_{mu}=G_{mu}(t):=g(t)-mu$.

From $k[f,g]=k[t]$ follows that, for all $lambda,mu in k$, $k[F_lambda,G_mu]=k[f-lambda,g-mu]=k[t]$

By the lemma we get that, for all $lambda,mu in k$,
$psi_{lambda,mu}: t mapsto (F_lambda(t),G_mu(t))$ is injective.

Assume that there exist $lambda_0,mu_0 in k$ such that
$deg(gcd(F_{lambda_0},G_{mu_0}))=deg(gcd(f-lambda_0,g-mu_0)) geq 2$.

Then there exist $a,b in k$, with $a neq b$ (otherwise, $a$ is a common root of $f'$ and $g'$ contrary to our assumption that they are not simultaneously zero) , such that $(t-a)(t-b)$ divides both $F_{lambda_0}$ and $G_{mu_0}$.

Then $F_{lambda_0}(a)=F_{lambda_0}(b)=G_{mu_0}(a)=G_{mu_0}(b)=0$,
so

$psi_{lambda_0,mu_0}(a)=(F_{lambda_0}(a),G_{mu_0}(a))=(F_{lambda_0}(b),G_{mu_0}(b))=psi_{lambda_0,mu_0}(b)$,

which contradicts the injectivity of
$psi_{lambda_0,mu_0}: t mapsto (F_{lambda_0}(t),G_{mu_0}(t))$.

Therefore, there exist no such $lambda_0,mu_0 in k$.

(ii) implies (i):

If $k[f,g] neq k[t]$, then by the lemma we get that $psi: t mapsto (f(t),g(t))$ is not injective, namely, there exist $a,b$, $a neq b$, such that $psi(a)=psi(b)$, so $f(a)=f(b):=lambda_0$ and $g(a)=g(b):=mu_0$

Take $F_{lambda_0}:=f-lambda_0$ and $G_{mu_0}:=g-mu_0$.

Then, $F_{lambda_0}(a)=f(a)-lambda_0=0$ and
$F_{lambda_0}(b)=f(b)-lambda_0=0$, so $(t-a)(t-b)$ divides $F_{lambda_0}$,
and $G_{mu_0}(a)=g(a)-mu_0=0$ and
$G_{mu_0}(b)=g(b)-mu_0=0$, so $(t-a)(t-b)$ divides $G_{mu_0}$.

We obtained that $(t-a)(t-b)$ divides $gcd(F_{lambda_0},G_{mu_0})=gcd(f-lambda_0,g-mu_0)$, so $deg(gcd(f-lambda_0,g-lambda_0)) geq 2$,
a contradiction.