show $\mathbb{Q}[t]/(t^2-1)\cong\mathbb{Q}\times\mathbb{Q}$

Question

I want to show $\mathbb{Q}[t]/(t^2-1)\cong\mathbb{Q}\times\mathbb{Q}$

My proof is as follows;
First, take $\varphi$ as follows \begin{array}{rccc} \varphi \colon &\mathbb{Q}[t] &\longrightarrow&\mathbb{Q}\times\mathbb{Q} \\ & f(t) & \longmapsto & (f(-1),f(1)) \end{array} then $\varphi$ is surjection and homomorphism, so $\mathbb{Q}[t]/\textrm{ker}(\varphi)\cong\mathbb{Q}\times\mathbb{Q}$ (from fundamental homomorphism theorem)
Next, show that $\textrm{ker}(\varphi)=(t^2-1)$.

(1)$\textrm{ker}(\varphi)\supset(t^2-1)$
$\forall f\in(t^2-1)\,,\,\,\varphi(f(t))=(0,0)$. so, $\textrm{ker}(\varphi)\supset(t^2-1)$

(2)$\textrm{ker}(\varphi)\subset(t^2-1)$
take $\forall f\in \textrm{ker}(\varphi)$. $f\in(t^2-1)$ when $f=0$. Then when $f\neq 0$, $f$ has factor $t-1$ since $f(1)=0$.　
In the same way, we can also see that we have $(t+1)$ as a factor.
Hence, $\textrm{ker}(\varphi)$ is ideal of $t^2-1=(t-1)(t+1)$.

I'm not very confident in this proof and would be happy to be corrected where wrong in this proof.
I appreciate your help.

Answer 1

Other than the typo mentioned in the comments, that it should be $\mathbb{Q}[t]/(t^2-1)$ and $\mathbb{Q}[t]/\ker\varphi$ rather than $\mathbb{Q}/(t^2-1)$ and $\mathbb{Q}/\ker\varphi$, nothing is actually wrong. The proof of $\ker(\varphi) \subset (t^2-1)$, however, can be made more formal/clear (it's not clear, for instance, why you split up into cases of $f=0$ and $f\neq 0$). Specifically, if you are checking whether $f$ is an element of $(t^2-1)$, the procedure that should come to mind is the division algorithm:

Suppose $f \in \ker(\varphi)$. By the division algorithm, there are $q, r \in \mathbb{Q}[t]$ such that $f = q\cdot (t^2-1)+r$ and $\deg r \leq \deg(t^2-1)-1 = 1$. Therefore there are $a,b \in \mathbb{Q}$ such that $$f = q\cdot (t^2-1) + at + b.$$ Apply $\varphi$ to find $$(0,0) = \varphi(f) = (q(-1)\cdot 0 + a(-1) + b, q(1)\cdot0 + a + b) = (b-a, b+a).$$ Then $b-a = 0$ implies $a=b$ and $$b=\frac{(b-a)+(b+a)}{2} = \frac{0+0}2 = 0.$$
It follows that $f = q\cdot (t^2-1) + 0 \in (t^2-1)$.