Prove that $2x$ is irreducible, but not prime in the ring of polynomials whose all coefficients, excepting the constant term, are even numbers.

Question

Let $$R=\{f \in \mathbb{Z}[x]:\text{ all coefficients of }f, \text{ except possibly the constant term, are even numbers}\}.$$ E.g., $1+6x+8x^2$ and $4+4x+10x^2$ are in $R$, but $2+5x$ is not in $R$. Prove that $R$ is an integral domain and $2x$ is an irreducible, but not prime, element of $R$.

I have tried to prove this by using cases such as $ab = 0$ if $a$ has constant even and $b$ constant odd and so on, then trying to show that either of $a$ or $b$ must be $0$ but I'm not getting it and I don't think that's the right approach. How can I show that $R$ is an integral domain? Also, how will I show that $2x$ is irreducible? Thank you.

Answer 1

Hint: for all of these problems, consider $R$ as a sub-ring of $\mathbb{Z}[x]$.

For the first part, recall that $\mathbb{Z}[x]$ is a integral domain – what can we conclude about subrings of integral domains?

For the second part, note that any factorization of $2x$ in $R$ will also be a factorization of $2x$ in $\mathbb{Z}[x]$, and recall that $\mathbb{Z}[x]$ is a unique factorization domain (thus in particular, the only factorizations of $2x$ in $\mathbb{Z}[x]$ up to units are the obvious ones). Can you finish this argument?

For the third part, note that every multiple of $2x$ in $\mathbb{Z}[x]$ will also be in the subring $R$. (Why?) So in particular, to come up with good candidates to witness the failure of $2x$ to be prime in $R$, think about those elements of $\mathbb{Z}[x]$ that are multiples of $2x$ in $\mathbb{Z}[x]$ but not in $R$. Example below:

For instance, $2x$ divides $4x^2=(2x)(2x)$ in $R$, but we also have that $4x^2=2\cdot(2x^2)$, both factors of which lie in $R$ but neither factor of which is a multiple of $2x$ in $R$.