Let I $=<x-2>$. Prove $\mathbb{Q[x]}/I \cong \mathbb{Q}$


Define $\phi: \mathbb{Q[x]} \rightarrow \mathbb{Q}$ by $\phi(f(x)) = f(2)$

I understand how to show that it is homomorphic and onto. I am struggling structuring the kernel.

$\textbf{Show onto:}$

$0\in \mathbb{Q} , \exists f(2) \in \mathbb{Q[x]} s.t. \phi(f(2)) = 0$

$\textbf{Show kernel:}$

Let $f(x) \in ker\phi$

$\phi(f(x)) = f(2)$


$a_0 + a_1(0) + a_2(0)^2 +.....+a_k(0)^k=0$

Hence $f(x) \in I$.

Thus $ker\phi \subseteq I$

Let $g(x) \in ker\phi$

$\phi(g(x)) = g(2)$

$b_0 + b_1(2) + b_2(2)^2 +.....+b_l(2)^l=0$

$\Rightarrow 2$ is a root of $g(x)$

Hence $g(x) \in I$.

Thus $I \subseteq ker\phi$

Thus $ker\phi = I$

By FHT, $\mathbb{Q[x]}/I \cong \mathbb{Q}$


Ok. First, if $f(x) \in \mathrm{ker}(\phi)$, then $\phi(f(x))=0$ so that $f(2)=0$. This means that $x=2$ is a root. Thus $(x-2)$ must be a factor of $f(x)$. So $f(x)=(x-2)g(x)$ for some $g(x) \in \mathbb{Q}[x]$. Since $f(x)$ is a multiple of $x-2$, we have $f(x) \in \langle x-2 \rangle$. [This shows $\mathrm{ker}(\phi) \subseteq \langle x-2 \rangle$.]

For the other direction, let $f(x) \in \langle x-2 \rangle$. Then $f(x)=(x-2)g(x)$ for some $g(x) \in \mathbb{Q}[x]$. Thus $f(2)=(2-2)g(2)=0$ so $\phi(f(x))=f(2)=0$. Therefore, $f(x) \in \mathrm{ker}(\phi)$. [This show $\langle x-2 \rangle \subseteq \mathrm{ker}(\phi)$.]

You have containment both ways, so done.

By the way, your "show onto" portion isn't a proof that $\phi$ is onto. To show $\phi$ is onto, suppose $r \in \mathbb{Q}$. Then the constant polynomial $f(x)=r$ belongs to $\mathbb{Q}[x]$ and $\phi(f(x))=f(2)=r$ (so $f(x)=r$ maps to $r$). Thus onto.


You may use the fact that if R is any integral domain, then given any polynomial f(X) in R[X] and a positive integer n, there exist polynomials r(X) and q(X) in R[X] such that, f(X) = X^n + a * q(X) + r(X), with r(X) = 0 or deg r(X) < n.

You may use this fact to prove the inclusion that ker /phi /subset I And I \subset ker \phi is very clear.


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