Radical Ideal for algebra Show the following:
a) $\text{rad}(IJ)=\text{rad}(I\cap J)=\text{rad}(I) \cap \text{rad}(J)$
b) $\text{rad}(I)=R$ if and only if $I=R$
c) if $P$ is prime $\text{rad}(P^n)= P$ for all $n$
d) Let $F$ be a field and $T$ a subset of $F^n$ . Show that the ideal
$I(T) = \{f ∈F[x_1,\ldots,x_n] : f(a_1,\ldots,a_n)=0$ for all $(a_1,\ldots,a_n) \in T\}$ is a radical ideal.
My answers so far:
a)Let $\text{rad}(I)=\{r \in R,r^n  \in I$,for some positive $n\}$ and $\text{rad}(J)=\{r \in R,r^m  \in J$, for some positive $m\}$. Then $r^nr^m = r^{n+m} \in IJ$. By homework 6 #4(b) we know that $IJ ⊆ I \cap J$ so $\text{rad}(IJ) \subseteq \text{rad}(I \cap J)$
Let $r \in \text{rad}(I\cap J)$ then $r^k \in (I \cap J)$ for some positive $k$. Let $k=n+m$ then $r^{n+m} \in (I \cap J)$ but we stated earlier that $r^{n+m} \in IJ$. Thus, $r \in  \text{rad}(IJ)$ and $r \in \text{rad}(I \cap J)$ so $\text{rad}(IJ) = \text{rad}(I \cap J)$.
Let $r\in \text{rad}(I \cap J)$ then $r^k \in I \cap J$ for some positive $k$.So $r^k  \in I$ and $r^k \in J$. Therefore $r \in \text{rad}(I) \cap \text{rad}(J)$ Let $r \in \text{rad}(I)$ and $r \in \text{rad}(J)$ then $r^k \in I$ and $r^k \in J$ for some positive $k$ then $r^k \in I \cap J$ so $r \in \text{rad}(I \cap J)$.
Thus, $\text{rad}(IJ) = \text{rad}(I \cap J) = \text{rad}(I) \cap \text{rad}(J)$. 
b)($\Longrightarrow$)Let $\text{rad}(I) = R$.
($\Longleftarrow$)Let $I = R$.
c) By part (a) we have $\text{rad}(P^n) = \text{rad}(P)$. Let $x \in P$ then $x^1 = x \in P$, so $x \in \text{rad}(P)$ showing that $P \subseteq \text{rad}(P)$. If $x \in \text{rad}(P)$ then $x^n \in P$ for some positive $n$. So $x^{nm} = (x^n)^m \in P$ for some positive $m$, then $x \in \text{rad}(P)$. Thus, $\text{rad}(P^n)=P$ for all $n$.
Need help with b and d.
 A: (1): $r(\mathfrak a \mathfrak b) = r(\mathfrak a \cap \mathfrak b) = r(\mathfrak a) \cap r(\mathfrak b)$:  
Proof: Notice that $\mathfrak a \mathfrak b \subseteq \mathfrak a \cap \mathfrak b$, and since the radical is monotone, $r(\mathfrak a \mathfrak b) \subseteq r(\mathfrak a \cap \mathfrak b)$. Let $x \in r(\mathfrak a \cap \mathfrak b)$, then there exists $n > 0$ such that $x^n \in \mathfrak a \cap \mathfrak b$. Hence $x^{2n} = x^nx^n \in \mathfrak a \mathfrak b$, so $x \in r(\mathfrak a \mathfrak b)$. Hence $r(\mathfrak a \cap \mathfrak b) \subseteq r(\mathfrak a \mathfrak b)$. Conclude that $r(\mathfrak a \mathfrak b) = r(\mathfrak a \cap \mathfrak b)$. 
Since $\mathfrak a \cap \mathfrak b \subseteq \mathfrak a, \mathfrak b$, and the radical is monotone, $r(\mathfrak a \cap \mathfrak b) \subseteq r(\mathfrak a), r(\mathfrak b)$ and so $r(\mathfrak a \cap \mathfrak b) \subseteq r(\mathfrak a) \cap r(\mathfrak b)$. For the reverse containment, let $x \in r(\mathfrak a) \cap r(\mathfrak b)$, so there exist $m, n > 0$ such that $x^m \in \mathfrak a$ and $x^n \in \mathfrak b$, therefore $x^{m + n} \in \mathfrak a \cap \mathfrak b$, conclude that $x \in r(\mathfrak a \cap \mathfrak b)$. Hence $r(\mathfrak a) \cap r(\mathfrak b) \subseteq r(\mathfrak a \cap \mathfrak b)$. Conclude that $r(\mathfrak a \cap \mathfrak b) = r(\mathfrak a) \cap r(\mathfrak b)$.
(2): $r(\mathfrak a) = (1)$ if and only if $\mathfrak a = (1)$: 
Proof: Suppose $r(\mathfrak a) = (1)$ and observe that $\mathfrak a \subseteq (1)$, so it suffices to show that $(1) \subseteq \mathfrak a$. Since $r(\mathfrak a) = (1)$, then there exists $n > 0$ such that $1^n = 1 \in \mathfrak a$. Conclude that $\mathfrak a = (1)$. 
Conversely, suppose that $\mathfrak a = (1)$ and observe that, $(1) \subseteq \mathfrak a \subseteq r(\mathfrak a) \subseteq (1)$ and thus $r(\mathfrak a) = (1)$.
(3): If $\mathfrak p$ is prime, $r(\mathfrak p^n) = \mathfrak p$ for all $n > 0$:
Proof: Let $\mathfrak p$ be prime and $n > 0$. We generalize part (1) via induction and get $r(\mathfrak p^n) = r(\mathfrak p)$. Observe that we have $\mathfrak p \subseteq r(\mathfrak p)$. For the reverse containment, let $x \in r(\mathfrak p)$, so there exists $m > 0$ such that $x^m \in \mathfrak p$, but since $\mathfrak p$ is prime, $x \in \mathfrak p$. Hence $r(\mathfrak p) \subseteq \mathfrak p$. Conclude that $r(\mathfrak p^n) = r(\mathfrak p) = \mathfrak p$.
(4): Let $k$ be a field and $T$ a subset of $k^n$. Show that the ideal $$I(T) = \{f \in k[x_1, \ldots, x_n] : f(a_1, \ldots, a_n) = 0 \text{ for all } (a_1, \ldots, a_n) \in T\}$$ is a radical ideal: 
Proof: It is immediate that $I(T) \subseteq r\big(I(T)\big)$, so it suffices to show $r\big(I(T)\big) \subseteq I(T)$. Let $f \in r\big(I(T)\big)$, then there exists $n > 0$ such that $f^n \in I(T)$ which means $$f^n(a_1, \ldots, a_n) = \big(f(a_1, \ldots, a_n)\big)^n = 0$$ for all $(a_1, \ldots, a_n) \in T$. Since $k[x]$ is an integral domain and $\big(f(a_1, \ldots, a_n)\big)^n = 0$ for all $(a_1, \ldots, a_n) \in T$, it follows that $f(a_1, \ldots, a_n) = 0$ for all $(a_1, \ldots, a_n) \in T$. Conclude that $f \in I(T)$ and hence we get the desired reverse containment.
