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A basic form of Hilbert's irreducibility theorem can be formulated as follows:

Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. There exist infinitely many linear polynomials $p(t)\in\mathbb{Q}[t]$, $p(t)=t-t_0$ such that $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$.

My question is: can we drop the "linear" assumption above and have polynomials $p(t)$ of arbitrarily high degree? More precisely, is the following true:

Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. Let $n$ be a positive integer. There exists a polynomial $p(t)\in\mathbb{Q}[t]$ such that $\deg p(t)>n$ and $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$.

Of course $p(t)$ must be irreducible itself, but that's far from enough.

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Let $L$ be a finite extension of ${\mathbb Q}$ over which $f(t,x)$ has absolutely irreducible factors, then for any $n$ there exists $K$ of degree $n$ over ${\mathbb Q}$ which is linearly disjoint from $L$, apply Hilbert Irreducibility over $K$ to $f(t,x)$ which is irreducible over $K$, so there exists $a \in K$ such that $K={\mathbb Q}(a)$ and $f(a,x)$ is irreducible over $K$, then $p(t)$ = minimal polynomial of $a$ over ${\mathbb Q}$ has degree $n$ and $(p(t), f(t,x))$ is a maximal ideal of ${\mathbb Q}[t,x]$.

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  • $\begingroup$ The problem is that every field contains $\mathbb{Q}$ hence no $K$ is going to be linearly disjoint from $L$. How could we force the element $a$ to be outside $\mathbb{Q}$? If the degree of extension is prime, just one element would do. $\endgroup$
    – Adam Przeździecki
    Commented Jul 8, 2018 at 5:49

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