Showing $\langle 2x+1,5\rangle$ is a maximal ideal

Question

Consider the integral domain $\mathbb{Z}[x]$. Show that the ideal $\langle 2x+1,5\rangle$ is a maximal ideal of $\mathbb{Z}[x]$.

I need help with this problem I don't even know where to start since its not 1 generated its throwing me off as to how I should proceed in solving it any hints or help is greatly appreciated thank you.

Answer 1

Notice that, for the Chinese Remainder Theorem, we have:$$\mathbb{Z[x]}/(2x+1,5)\cong\mathbb{Z_5[x]}/(2x+1)$$

Now we know that if R is an integral domain and I is a maximal ideal, R/I is a field.

$(2x+1)$ is a maximal ideal $\iff$ $2x+1$ is irreducible over $\mathbb{Z_5[x]}$, which it is. So you have your thesis.

Answer 2

Here is one direct proof. Note that $6x+3$ is in this ideal and therefore $x+3$ is in the ideal. It follows that any polynomial $p(x)\in\mathbb{Z}[x]$ is congruent to an integer module the ideal. Now if we add any element to this ideal that is the same as adding an integer. If we add an integer not divisible by $5$, it will generate $\mathbb{Z}[x]$ and thus the ideal is maximal.