I'm just learning about ideal and struggling with these problem. So I want to know how to prove these.
About ℚ[x] ⊂ C[X], a ∈ C,
$I_a = \{f(X) ∈ ℚ[x] | f(a) = 0\}$
(1) Show $I_a$ is an ideal of $ℚ[x]$
(2) show $I_{2^{1/2}} = \left\{(X^2-2)g(X) | g(X) ∈ ℚ[x]\right\}$
(you can use $2^{1/2}$ is irrational number without condition)
To solve this problem, can I use under these conditions?
(a) I ≠ ø
(b) f(X), g(X) ∈ I, f(X)+ g(X) ∈ I
(c) f(X) ∈ I and a(X)∈ K[X], f(X)a(X) ∈ I
In addition, to prove problem(1) using (a), is it correct to use constant(the element of ℚ)is included in ℚ[x]?
+) To solve problem(2), I tried to use GCD(f(x), (x^2-2)) = GCD((x^2-2), r(x))
so, I have to consider when r(x)=0 and deg(r(x)) < 2
when r(x)= 0,
GCD(f(x), (x^2-2)) = GCD((x^2-2), 0)
GCD(f(x), (x^2-2)) = (x^2-2)
however, I got stuck after this.
1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) \in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) \in I_a$
Let $f(x) \in I_a$ and $p(x) \in \mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) \in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) \in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) \in \mathbb{Q}[x]$ we have $f(x)+g(x) \in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) \in I_{2^{1/2}}$ and $p(x) \in \mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) \in \mathbb{Q}[x]$ we have $f(x)p(x) \in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.