Distinct elements of $\mathbb{Z}[x]/I$ Write two distinct elements of $\mathbb Z[x] / \langle x-7, 13\rangle$.
 A: Hint $\rm\ 1\not\equiv 0\pmod I\:$ else $\rm\:1\in I\ \Rightarrow\ 1\: =\: 13\:f(x) + (x-7)\:g(x)\ \Rightarrow\ 1 =  13\:f(7) \:$ in $\mathbb Z$
Remark $\ $ This shows that $\rm\:\mathbb Z\to \mathbb Z[x]\to \mathbb Z[x]/I\:$ has kernel $\rm\:13\!\:\mathbb Z.\:$ Further, this map is onto since $\rm\:f(x)\equiv f(7)\pmod{I}.\:$ Hence $\rm\:\mathbb Z[x]/I\:\!\cong\!\: \mathbb Z/13\:$ by the First Isomorphism Theorem.
A: What is $x$ in this ring? Once you know that, you should have a good idea what the ring looks like, and from there the problem is easy.
Here's a cheap answer (that you should NOT use as your solution): The problem asks for two distinct elements of the ring. Therefore, (for completely non-mathematical reasons) we assume that the ring has 2 distinct elements. We know $1$ is in the ring, and so is $0$, and $1 \neq 0$ in all rings with unit except the ring with one element.
A: Let $\pi : \mathbb{Z}[x] \to \mathbb{Z}[x]/I$ be the canonical projection. We claim that $\pi(1)$ and $\pi(2)$ are distinct elements in $\mathbb{Z}[x]/I$. Otherwise, $2 - 1 = 1 \in I$, and $I$ would be the unit ideal. Now, show that $I$ is not, in fact, the unit ideal.
