Justification of an isomorphism between $\mathbb{Z}[t]/(t,3)$ and $\mathbb{Z}/3$ I am wondering if the following is true:
$$\mathbb{Z}[t]/(t,3)\cong\mathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $\mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $\mathbb{Z}[t]/(t,3)\cong(\mathbb{Z}[t]/t)/(3)\cong\mathbb{Z}/3$, where the $(3)$ is an ideal in $\mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
 A: Yes your intuition is correct and I try to explain necessary steps:

Claim. $\Bbb{Z}[x]/\langle x,3\rangle\cong\Bbb{Z}/\langle3\rangle$

Well, I'll prove the following steps:
$$\Bbb{Z}[x]/\langle x,3\rangle\cong \frac{\Bbb{Z}[x]/\langle x\rangle}{\langle x,3\rangle/\langle x\rangle} \cong\Bbb{Z}/\langle 3\rangle$$


*

*The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.

*I think $\Bbb{Z}[x]/\langle x\rangle \cong \Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) \mapsto f(0)$ form $\Bbb{Z}[x]$ to $\Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.

*Now for the equality $\langle x,3\rangle/\langle x\rangle\cong \langle 3\rangle$ Try to define the homomorphism as follows:
$\phi :\langle x,3\rangle \to \langle 3\rangle$ by mapping any element $xf(x)+3g(x)\in \langle x,3 \rangle$ to $3g(0) \in \langle 3\rangle$ (N.B. Here, $f,g \in \Bbb{Z}[x]$), i.e. $$\phi: xf(x)+3g(x) \mapsto 3g(0)$$
See that this is an epimorphism. Also $\ker \phi=\langle x\rangle$. Then by 1st Isomorphism theorem $\langle x,3\rangle/\ker \phi \cong \langle 3\rangle$. Which justifies all your steps...!!
N.B. To find out the $\ker \phi$, remember for any $g(x)\in \Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $g\in \langle x\rangle$.
A: It's legitimate and a consequence of the isomorphism theorems. 
