Let $D$ be an integral domain and let $c\in D$ be irreducible in $D$. Show the ideal $(x,c)$ in $D[x]$ is not principal. [duplicate]

Hint: if $R$ is a domain and not a field, take $r\in R$ such that $r\ne0$ and $r$ is not invertible. Then the ideal $(r,x)$ in $R[x]$ is not principal.

Hint $ $ If $\,(c,x)=(f)\,$ then $\,f\mid c\overset{\rm\color{#90f}{domain}}\Rightarrow \deg f = 0,\,$ so $\,\color{#0a0}{f(1)=f(0)},\,$ so eval $\,(c,x)=(f)\,$ at $\,x=0\Rightarrow \color{#c00}{(c)}=\color{#0a0}{(f(0))},\,$ and eval at $\,x=1\Rightarrow (1)= (\color{#0a0}{f(1)}) = (\color{#0a0}{f(0)}) = \color{#c00}{(c)}\Rightarrow \color{#c00}c\mid 1\,$ (idea behind proof of Lemma below). $ $ Apply the Lemma with $\,\rm C = K[y],\,$ where $\rm \,0\neq y\,$ is not a unit.

Lemma $ $ If $\ \rm \color{#90f}{0\neq c\in domain}\ C\ $ then $\rm \ (c,x) = (f)\ $ in $\rm \,C[x]\,\Rightarrow \,c\,$ is a unit in $\,\rm C.\ $ Proof:

$\rm\ \ f\ \in\ (c,x)\ \Rightarrow\ f = c\, g_1 + x\,h_1.\, $ Eval at $\rm\: x \!=\! 0\ \Rightarrow\ \color{#0a0}{f(0)} = cg_1(0) = \color{#0a0}{cd},\,\ d\in C$

$\rm\ \ c\ \in\ (f)\ \Rightarrow\ c\ =\ f\, g, \ \,\color{#90f}{hence}\,\ \ \deg f\:\! =\:\! 0\:\Rightarrow\: \color{#c00}f = \color{#0a0}{f(0) = cd}$

$\rm\ \ x\ \in\ (f)\ \Rightarrow\, x\ =\ \color{#c00}f\, h\ =\ \color{#0a0}{cd}\:\!h.\, $ Eval at $\rm\ x\! =\! 1\ \Rightarrow\ 1 = cd\,h(1)\ \Rightarrow\ c\,$ is a unit in $\rm\,C$

Remark $ $ See here for general characterizations of when a polynomial ring or semigroup ring $R[S]$ is a PIR (principal ideal ring).

Consider the function $\phi\colon K[x,y]\to K$ defined by $\phi(f) = f(0,0)$ (that is evaluation at $(0,0)$). Check that it is a homomorphism.

So its kernel consists of polynomials of two variables vanishing at origin; that is they have no constant terms. This is not a principal ideal (in fact all maximal ideals of this ring are non principal).