Exercise 3.1.3

Proof. Let $a,b\in A=F[x]$. Suppose that $a+I=a^{\prime }+I$ and $b+I=b^{\prime }+I$.

Then $a^{\prime }=a+u,u\in I,b^{\prime }=b+v,v\in I$, so $a^{\prime }+b^{\prime }=a+b+u+v$, where $u+v\in I$, thus $a+b+I=a^{\prime }+b^{\prime }+I$.

$a^{\prime }{b}^{\prime }=\mathit{ab}+\mathit{bu}+\mathit{av}+\mathit{uv}$, where $\mathit{bu}+\mathit{av}+\mathit{uv}\in I$, so $\mathit{ab}+I=a^{\prime }{b}^{\prime }+I$.

The equivalence relation $\sim$ defined on $A$ as $a\sim a^{\prime }⟺a+I=a^{\prime }+I(⟺a^{\prime }-a\in I)$ is so compatible with addition and multiplication in $A$, and the class of an element $a\in A$ is $a+I$. We can so define sum and product of two classes by

If $\phi :A\to A∕I$ is defined by $\phi (a)=a+I$, then (1) and (2) are written

Moreover $\phi (1)=1+I$ is the multiplicative identity of $A∕I$.

$\phi :A\to A∕I$ is a ring homomorphism. □