I've to solve the following problem.

$R=\mathbb Q [x]$ and for $S \subset \mathbb C$ define $I(S):=\{f \in \mathbb Q [x]: f(x)=0 \forall x \in S \}$.

I have already shown, that this is an ideal in $\mathbb Q [x]$.

The question is now:

For which $S_i$ is $I(S_i)$ a prime ideal ?

$S_1 := \{3,7\}$, $S_2 := \{7+3i,7-3i\}$, $S_3 := \{\sqrt[3]2,-\sqrt[3]2\}$,$ S_4 := \{\sqrt2,-\sqrt2\}$

For $S_1$: it is a prime ideal.

$I(S_1) \neq R$, because of the polynomial $f=x$ with $f(3)=3 \neq 0$ and $f(7)=7 \neq 0$.

Let $a,b \in R$ such that $ab \in I(S_1)$, so $(ab)(x)=0 \forall x \in S_1$. Now I know that $0=(ab)(x)=a(x)b(x) \forall x \in S_1$. This leads to $a(x)=0$ or $b(x)=0$ , because there are no zero divisors in $\mathbb Q[x]$. So $a \in I(S_1)$ or $b \in I(S_1)$, which is the definition of prime ideal.

So now I need the other $S_i$. I hope someone of you can help me.

Edit: This "proof" is wrong, because of the polynomials $a(x)=x-3$ and $b(x)=x-7$.

  • 2
    $\begingroup$ @Myrkuls JayKay: Your argument for $S_1$ is flawed. Consider $a,b$ given by $a(x) = x-3$ and $b(x) = x-7$. $\endgroup$ – quasi May 25 '17 at 19:12
  • $\begingroup$ It might help to note that non-zero prime ideal = maximal ideal in $\mathbb{Q}[x]$. $\endgroup$ – Steve D May 25 '17 at 19:20

Hints: Argue that \begin{align*} I(S_1)&=((x-3)(x-7))\\[4pt] I(S_2)&=(x^2-14x+58)\\[4pt] I(S_3)&=((x^3-2)(x^3+2))\\[4pt] I(S_4)&=(x^2-2)\\[4pt] \end{align*} and recall that an ideal $(f)$ of $\mathbb{Q}[x]$ is a prime ideal if and only if $\deg(f) \ge 1$ and $f$ is irreducible in $\mathbb{Q}[x]$.

  • $\begingroup$ Thank you, but I dont know that fact. Is it easy to proof ? $\endgroup$ – Myrkuls JayKay May 25 '17 at 20:00
  • 1
    $\begingroup$ One direction is immediate: Let $f \in \mathbb{Q}[x]$ with $\deg(f)>1$, and suppose $f=gh$ for some $g,h \in \mathbb{Q}[x]$ with $\deg(g) < \deg(f)$ and $\deg(g)<\deg(f)$. Then $gh \in (f)$ but $g,h \notin (f)$ (since all elements of $(f)$ have degree at least that of $f$). Thus, if $f$ is reducible in $\mathbb{Q}[x]$, the ideal $(f)$ is not a prime ideal of $\mathbb{Q}[x]$. $\endgroup$ – quasi May 25 '17 at 20:14
  • 1
    $\begingroup$ For the other direction, let $f \in \mathbb{Q}[x]$ with $\deg(f)>1$ be irreducible in $Q[x]$, and suppose $gh \in (f)$ for some nonzero $g,h \in \mathbb{Q}[x]$. Then $gh=af$ for some $a \in \mathbb{Q}[x]$. Since $\mathbb{Q}[x]$ is a UFD, $gh$ factors uniquely (apart from order and unit [nonzero constant] factors) into irreducible factors. Since $f$ is irreducible, it must be one of the irreducible factors of $gh$, hence (by uniqueness) must be one of the irreducible factors of $g$ or $h$. It follows that at least one of $g,h$ is in $(f)$. Therefore $(f)$ is a prime ideal of $\mathbb{Q}[x]$. $\endgroup$ – quasi May 25 '17 at 20:30

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