Which of the sets are ideals and maximal ideals?

Question

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?

Answer 1

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.

Answer 2

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 $\boldsymbol{c)}$:

$J=(x-\sqrt 3)\mathbf R[x]\cap \mathbf Q[x]$.