Question

If $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$ is a surjective homomorphism of integral domains, p is irreducible in R, and$\mathit{f}\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{\ne }{\mathbf{0}}_{\mathbf{S}}$ is$\mathit{f}\mathbf{\left(}\mathbf{p}\mathbf{\right)}$ irreducible in S?

Short answer

No,$f\left(p\right)$ is not irreducible in$S$.

Step-by-step solution

Define a map f:ℤ[x]→ℤ defined as  f(x)=f(3)

Now, this is an evaluation function and it is a homomorphism.

It is surjective since the constant function in$\mathbb{Z}\left[x\right]$ are mapped to themselves.

Take a function f(x)=x2+1∈ℤ[x]

It is irreducible.

Notice that $f(x^2+1)=3^2+1=10$ and 10 is not irreducible in $\mathrm{\mathbb{Z}}$.

Therefore,role="math" localid="1654614962835" $f\left(p\right)$ is not irreducible in $S$.