I am not sure exactly what you have in mind. But

(1) If your polynomial $P \in \mathbb Q(x)$, and $K%$ is some field containing $K$, then obviously it will stay reducible over $K$.

(2) You can assume that your polynomial is monic, and then its coefficients are in $\mathbb Z[1/N]$ for some positive integer $N$, so you can reduce your polynomial mod $p$ for all primes $p$ not dividing $N$. Unfortunately, and obviously, they polynomials you get in $\mathbb F_p[x]$ this way will still be reducible.

So there is no way to make it irreducible by reduction/extension. However, depending on what you want to do after, there is something that you can do:

(3) Put your polynomials $P(x)$ in a family of polynomials $P_t(x)$ depending on some parameter $t \in \mathbb Q$, such that $P_0(X)=P(x)$. It easy to choose the family such that $P_t(x)$ becomes irreducible on $\mathbb Q(x,t)$, and this will imply that $P_t(x)$ will be irreducible over $\mathbb Q(x)$ for all values of $t$ except a (edited:) small set (including your starting point $t=0$) by Hilbert's Irreducibility Theorem (see e.g. wikipedia). Hence, you may be able to prove things for the irreducible polynomials in your family, and deduce similar things by "continuity" for your initial $P_0(X)$.

I am not sure exactly what you have in mind. But

(1) If your polynomial $P \in \mathbb Q(x)$, and $K%$ is some field containing $\mathbb{Q}$, then obviously it will stay reducible over $K$.

(2) You can assume that your polynomial is monic, and then its coefficients are in $\mathbb Z[1/N]$ for some positive integer $N$, so you can reduce your polynomial mod $p$ for all primes $p$ not dividing $N$. Unfortunately, and obviously, they polynomials you get in $\mathbb F_p[x]$ this way will still be reducible.

So there is no way to make it irreducible by reduction/extension. However, depending on what you want to do after, there is something that you can do:

(3) Put your polynomials $P(x)$ in a family of polynomials $P_t(x)$ depending on some parameter $t \in \mathbb Q$, such that $P_0(X)=P(x)$. It easy to choose the family such that $P_t(x)$ becomes irreducible on $\mathbb Q(x,t)$, and this will imply that $P_t(x)$ will be irreducible over $\mathbb Q(x)$ for all values of $t$ except a (edited:) small set (including your starting point $t=0$) by Hilbert's Irreducibility Theorem (see e.g. wikipedia). Hence, you may be able to prove things for the irreducible polynomials in your family, and deduce similar things by "continuity" for your initial $P_0(X)$.

I am not sure exactly what you have in mind. But

(1) If your polynomial $P \in \mathbb Q(x)$, and $K%$ is some field containing $K$, then obviously it will stay reducible over $K$.

(2) You can assume that your polynomial is monic, and then its coefficients are in $\mathbb Z[1/N]$ for some positive integer $N$, so you can reduce your polynomial mod $p$ for all primes $p$ not dividing $N$. Unfortunately, and obviously, they polynomials you get in $\mathbb F_p[x]$ this way will still be reducible.

So there is no way to make it irreducible by reduction/extension. However, depending on what you want to do after, there is something that you can do:

(3) Put your polynomials $P(x)$ in a family of polynomials $P_t(x)$ depending on some parameter $t \in \mathbb Q$, such that $P_0(X)=P(x)$. It easy to choose the family such that $P_t(x)$ becomes irreducible on $\mathbb Q(x,t)$, and this will imply that $P_t(x)$ will be irreducible over $\mathbb Q(x)$ for all values of $t$ except a finite set (including your starting point $t=0$) by Hilbert's Irreducibility Theorem (see e.g. wikipedia). Hence, you may be able to prove things for the irreducible polynomials in your family, and deduce similar things by "continuity" for your initial $P_0(X)$.

I am not sure exactly what you have in mind. But

(1) If your polynomial $P \in \mathbb Q(x)$, and $K%$ is some field containing $K$, then obviously it will stay reducible over $K$.

(2) You can assume that your polynomial is monic, and then its coefficients are in $\mathbb Z[1/N]$ for some positive integer $N$, so you can reduce your polynomial mod $p$ for all primes $p$ not dividing $N$. Unfortunately, and obviously, they polynomials you get in $\mathbb F_p[x]$ this way will still be reducible.

So there is no way to make it irreducible by reduction/extension. However, depending on what you want to do after, there is something that you can do:

(3) Put your polynomials $P(x)$ in a family of polynomials $P_t(x)$ depending on some parameter $t \in \mathbb Q$, such that $P_0(X)=P(x)$. It easy to choose the family such that $P_t(x)$ becomes irreducible on $\mathbb Q(x,t)$, and this will imply that $P_t(x)$ will be irreducible over $\mathbb Q(x)$ for all values of $t$ except a (edited:) small set (including your starting point $t=0$) by Hilbert's Irreducibility Theorem (see e.g. wikipedia). Hence, you may be able to prove things for the irreducible polynomials in your family, and deduce similar things by "continuity" for your initial $P_0(X)$.