Which of the sets are ideals and maximal ideals? The exercise asks me to prove which of the sets are ideals, and if they are, which of those are maximal. 
I have these 4 cases:
$$
a) J = \{f(x)\in \mathbb{Q}[x]: f(1)=f(7)=0 \}
\\b) J = \{f(x)\in \mathbb{Q}[x]: f(2)=0;f(5)\neq 0 \}
\\c) J = \{f(x)\in \mathbb{Q}[x]: f(\sqrt{3})=0 \}
\\d) J = \{f(x)\in \mathbb{Q}[x]: f(4)=0; f(0)=1 \}
$$
At least for the case $c$, I think I just have to prove that this ideal is generated by $x^2-3$ and them, since this polynomial is irreducible over $\mathbb{Q}[x]$, we have that it's a maximal ideal. I'm basing myself on my other question here.
For $a$, I know that if a polynomial is $0$ in $x=1, x=7$, whenever I multiply it by another polynomial, let's say $g(x)$, we'd have:
$$g(1)f(1) = g(1)\cdot 0 = 0\\g(7)f(7) = g(7)\cdot 0 = 0$$
thus this is an ideal, but how do I verify if it is maximal or not?
Now, for $b$, when we multiply a polynomial by $f(2)$, our new polynomial will still be $0$ at $x=2$, but for $x=5$ it may be $0$, for example, multiply the polynomial $p(x) = x-5$ by $f(x)$ and evaluate it at $x=5$, we have:
$$0\cdot f(5)= 0$$
Therefore this is not an ideal.
Now, case $d$ gives me some doubts. Could you guys help me?
 A: For $a)$  remember $f(\alpha)=0\iff f(x) $ is a multiple of $x-\alpha$. Hence in this case $f(x)\in (x-1)\mathbf Q[x]\cap (x-7)\mathbf Q[x]$, which is none other than $(x-1)(x-7)\mathbf Q[x]$ Conversely any polynomial in this ideal has $1$ and $7$ as roots.
It is not a maximal ideal since it is not generated by an irreducible polynomial.
Cas $d)$: if it were an ideal,it would imply that any multiple of a polynomial with constant term $1$ also has constant term $1$, which is absurd.
Hint for question $\boldsymbol{c)}$:
$J=(x-\sqrt 3)\mathbf R[x]\cap \mathbf Q[x]$.
A: The sets in b and d are not ideals, because they don't contain the zero polynomial.
The sets in a and c are ideals and this can be proved directly.

Lemma. If $\alpha\in\mathbb{C}$, then $I_\alpha=\{f(x)\in\mathbb{Q}[x]:f(\alpha)=0\}$ is an ideal of $\mathbb{Q}[x]$.
Proof. (1) The zero polynomial belongs to $I_\alpha$, which is obvious.
(2) If $f,g\in I_\alpha$, then $f+g\in I_\alpha$. Indeed…
(3) If $f\in I_\alpha$ and $g\in\mathbb{Q}[x]$, then $fg\in I_\alpha$. Indeed…

(Fill in the details.)
Now the set in a is $I_1\cap I_7$, so it's the intersection of two ideals. It is not maximal, because $I_1\cap I_7\subsetneq I_1$: indeed, $x-1\in I_1$, but $x-1\notin I_1\cap I_7$. Moreover $I_1$ is a proper ideal, because it doesn't contain the nonzero constant polynomials. In other words, $I_1\cap I_7\subsetneq I_1\subsetneq \mathbb{Q}[x]$.
The set in d is $I_{\sqrt{3}}$, so it is an ideal.
Let $f(x)\in I_{\sqrt{3}}$; then by the division algorithm, we have
$$
f(x)=(x^2-3)Q(x)+R(x)
$$
in $\mathbb{Q}[x]$, where $R(x)=0$ or $R(x)$ has degree less than $2$. But no degree $0$ or $1$ polynomial with rational coefficients has $\sqrt{3}$ as a root. Therefore $R(x)=0$ and so $f(x)=(x^2-3)Q(x)$. Conversely, every polynomial of the form $(x^2-3)Q(x)$ vanishes at $\sqrt{3}$. Hence
$$
I_{\sqrt{3}}=\langle x^2-3\rangle
$$
Since $x^2-3$ is an irreducible polynomial in $\mathbb{Q}[x]$, which is a principal ideal domain, the ideal generated by $x^2-3$ is maximal.
