$\mathbb{Z}[X]/(X,3) \cong \mathbb{F}_3$

Question

I am trying to understand why $\mathbb{Z}[X]/(X,3) \cong \mathbb{F}_3$ is true.

I tried to do the following but I got stuck: $\mathbb{Z}[X]/(X,3)\cong (\mathbb{Z}[X]/X)/((X,3)/(X))\cong ???$

Could someone please help me? Thank you

Answer 1

Hint: Define an epimorphism $\mathbf Z[X]\to \mathbf F_3$ whose kernel is $(X,3)$.

Answer 2

A more intuitive way to see this: We start by $\mathbb{Z}[x]$. If we "cut" by $x$, all we are left with is $\mathbb{Z}$. If we cut by multiples of $3$, we are left with $\mathbb{Z}_3$.

A more accurate way to see this: Define $\varphi:\mathbb{Z}[x]\to\mathbb{Z}$ by $\varphi(f)=f(0)$. This is an epimorphism with kernel $\langle x\rangle$. Thus $\mathbb{Z}\cong\mathbb{Z}[x]/\langle x\rangle$, by the 1st isomorphism theorem. Now compose the epimorphism $\psi:\mathbb{Z}\to\mathbb{Z}_3$ given by $\psi(n)=[n]_3$ with the previous isomorphism. The 1st isomorphism theorem will yield the result.

Answer 3

Hint:

The ideal $\:3\cdot\mathbf Z[X]/(X)=(3,X)/(X)$, so by the third isomorphism theorem, $$\mathbf Z[X]/(3,X)\simeq\mathbf Z[X]/(X)\Big/(3,X)/(X)$$