In $\mathbb{Z}[t]$, $Q = (4, t)$ is not a power of $M = (2, t)$

Question

The problem of showing that Q, as above, is not a power of M, as above, rises as part of a larger problem. I'm confident about my response to the other parts, but the best justification I can come up with for this particular fact is that $(2, t)^n$ will always have $2t^{n-1}$ as a generator (where $n = 2, 3, ...$), and thus is never a subset of, much less equal to, $Q$. But this feels shaky--I think it's true, but don't know how to prove it. Is there a way to be certain of this statement, or a better way to show that $Q$ can't possibly be a power of $M$?

Answer 1

The $n$th power of $(2,t)$ is $(2^n,2^{n-1}t,2^{n-2}t^2,\dots,2t^{n-1},t^n)$. It follows immediately that every element of this power is a polynomial whose coefficient of $t$ is divisible by $2^{n-1}$.

No such power, then, contains $t$.

Answer 2

Hint $\,\ \Bbb Z[t]/(2,t)^n \cong \Bbb Z[t]/(4,t)\cong \Bbb Z/4\,\Rightarrow\,n=1\,\Rightarrow 2 = 4\,\Rightarrow\!\Leftarrow$

Alternatively $\ {\rm mod}\ 2\!:\ (2,t)^n = (4,t)\,\Rightarrow\,(t^n) = (t)\,\Rightarrow\, n=1\,\overset{\large {\rm mod}\ t}\Rightarrow\, (2)=(4)\,$ in $\,\Bbb Z \,\Rightarrow\!\Leftarrow$