Let \(I\) be the ideal of \(\mathbb Z[x]\) generated by \(x-7\) and \(15\). Prove that \[ \mathbb Z[x]/I\cong\mathbb Z/15\mathbb Z. \]
Solution
1. Define\[ \varphi:\ZZ[x]\longrightarrow\ZZ/15\ZZ, \qquad \varphi(p)=p(7)\pmod{15}. \]Then \(\varphi\) is a surjective ring homomorphism.
Evaluation at \(7\) is a ring homomorphism \(\ZZ[x]\to\ZZ\), and reduction modulo \(15\) is a ring homomorphism \(\ZZ\to\ZZ/15\ZZ\), so their composite is a ring homomorphism. It is surjective because every residue class is the image of a constant polynomial.
2. The ideal\[ I=(x-7,15) \]is contained in \(\ker\varphi\).
One has \[ \varphi(x-7)=0 \qquad\text{and}\qquad \varphi(15)=0. \] Therefore the ideal generated by these two elements lies in the kernel.
3. One has\[ \ker\varphi\subseteq I. \]
Let \(p\in\ker\varphi\). Then \[ p(7)=15m \] for some \(m\in\ZZ\). By the polynomial remainder theorem, \[ p(x)-p(7)=(x-7)q(x) \] for some \(q(x)\in\ZZ[x]\). Hence \[ p(x)=(x-7)q(x)+15m\in(x-7,15)=I. \]
4. Thus\[ \ker\varphi=I. \]
Combine steps 2 and 3.
5. Therefore\[ \boxed{\ZZ[x]/I\cong\ZZ/15\ZZ}. \]
By step 1, \(\varphi\) is surjective, and by step 4 its kernel is \(I\). The first isomorphism theorem for rings gives the result.
6. Q.E.D.
Step 5 is the required isomorphism.