1. ## Irreducible polynomials

Suppose $\displaystyle g(x)$ is an irreducible in R. If $\displaystyle f(x)$ is any other polynomial, does that imply that $\displaystyle g(f(x))$ is also irreducible?

I'm particularily interested in the case when $\displaystyle g(x) = x^2 + c$.

Here are some examples I've tried

$\displaystyle g(x) = x^2 + d$

1) $\displaystyle f(x) = c$

$\displaystyle g(f(x)) = c^2 + d$ is a constant

2) $\displaystyle f(x) = x + c$

$\displaystyle g(f(x)) = x^{2} + 2c \cdot x + c^2 + d$

So if I choose (c,d) such that the above has integer roots then I've solved the problem.

So I need $\displaystyle 4c^2 - 4 \cdot (c^2 + d)$ to be a square. But $\displaystyle 4$ divides this, so I just need for $\displaystyle c^2 - c^2 - d = -d$ to be a square.

This is a contradiction to the condition that $\displaystyle g(x) = x^2 + d$ be irreducible because then g(x) would then be a difference of squares.

I stopped after this......Naturally the next f(x) to try would be a quadratic, but I don't know enough about quartic polynomials to know the conditions for when they're irreducible or not.

2. I assume from the way that the problem is phrased that the scalars here are the real numbers. If so, let $\displaystyle g(x) = x^2+1$ (irreducible) and let $\displaystyle f(x) = x^2$. Then $\displaystyle g(f(x)) = x^4+1 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$ (reducible).

With a bit more effort, you can get a similar example with rational scalars.

3. Thanks. Thats a good start and got me thinking along similar lines.

Using $\displaystyle g(x) = x^2 + 1$ and $\displaystyle f(x) = x^d$ where $\displaystyle d$ is odd. would give a polynomial $\displaystyle x^{2d} + 1 = (x^2 + 1) \cdot (x^{2d-2} - x^{2d-4} + x^{2d-6} - ..... + 1)$.