Consider the $\mathbb Q$-algebra homomorphism $\varepsilon_{\sqrt 2}:\mathbb Q[X]\rightarrow \mathbb C$ defined by $\varepsilon(X)=\sqrt 2$.
How to determine the kernel $\ker \varepsilon_{\sqrt 2}$ and the image $\varepsilon_{\sqrt 2}(\mathbb Q[X])?$
You can think of this homomorphism as essentially evaluating any polynomial in $Q[X]$ at the point $\sqrt{2}$, as it sends $g(x)$ to $g(\sqrt{2})$. (here we are utilizing the homomorphism properties of this particular map)
To find the kernel, you need to find which polynomials get sent to 0 if you evaluate them at the point $\sqrt{2}$. $X^2-2$ is a prime example of such a polynomial, so at this point you might have realised that any multiple of $X^2-2$ vanishes when evaluated at $\sqrt{2}$. So, the kernel is the ideal $(X^2-2)$.
Proof: suppose that $p(x) \in Ker(\epsilon_{\sqrt2})$. Then using euclidean division, you may write $p(x)=(x^2-2)f(x)+r(x)$, and this means that $0=p({\sqrt{2}})=r(\sqrt{2}).$
$r(x)$ is a polynomial of degree at most 1, but there's no polynomial of degree 1 over the rationals that sends $\sqrt{2}$ to zero, apart from the zero polynomial.
This means that r=0, and therefore any polynomial in $Ker(\epsilon_{\sqrt2})$ is a multiple of $(X^2-2)$.
For $Im(\epsilon_{\sqrt2})$, you may start with the definition of the an arbitrary element in $Q[x]$, which looks like $\sum_{i=0}^{\infty}a_ix^{i}$ but only finitely many $a_i$ are non-zero. The action of this map simply sends X to $\sqrt{2}$, ie you get $Q(\sqrt{2})$