Intersection of Ideals and Radical Membership

For $I,J \in k\left[\textbf{x}\right]$ prove that

$\left( a \right)$

\begin{align} I\cap J = \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right] \end{align}

$\left( i \right)$ $I\cap J \subset \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right].$
Suppose $f \in I\cap J$. So $f \in I$ and $tf \in tI$. Similarly, $f \in J$ and $\left(1-t \right)f \in \left(1-t \right)J$. Thus $f = tf + \left( 1-t \right) \in \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]}.$ Additionally, $I,J \in k\left[\textbf{x}\right]$. Hence, $f \in \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right]$ and $I\cap J \subset \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right].$

$\left( ii \right)$ $\left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right] \subset I\cap J.$
Suppose $f \in \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]}.$ Then we can write $f\left( x \right)$ as $f\left( x \right) = g\left( x,t \right) + h\left( x,t \right)$ where $g\left( x,t \right) \in tI$ and $h\left( x,t \right) \in \left( 1-t \right)J$. If $t = 0$, then $f \in J \cap k\left[\textbf{x}\right] = J$. If $t = 1$, then $f \in I \cap k\left[\textbf{x}\right] = I$. So $f \in I \cap J$ and $\left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right] \subset I\cap J$.

Therefore, $I\cap J = \left\langle tI+\left(1-t\right)J\right\rangle_{k\left[\textbf{x},t\right]} \cap k\left[\textbf{x}\right]$.

$\left( b \right)$

\begin{align} f \in \sqrt{I} \Leftrightarrow 1 \in \left\langle I, tf - 1\right\rangle_{k\left[\textbf{x},t\right]} \end{align}

$\left( \Leftarrow \right)$ If $1 \in \left\langle I, tf-1\right\rangle_{k\left[\textbf{x},t\right]}$, then Hilbert's Nullstellensatz implies that $f^{n} \in I$ for some positive integer $n$. This, in turn, implies that $f \in \sqrt{I}$.

$\left( \Rightarrow \right)$ If $f \in \sqrt{I}$ then $f^{n} \in I \subset \left\langle I, tf-1 \right\rangle_{k\left[\textbf{x},t\right]}$ for some $n$ and $tf-1 \in \left\langle I, tf-1 \right\rangle_{k\left[\textbf{x},t\right]}$. Rewriting $1$ gives us

\begin{align} 1 = t^{n}f^{n} + \left(1 - t^{n}f^{n}\right) = t^{n}f^{n} + \left(1 - tf\right)\cdot\left(1 + tf + \cdots + t^{n-1}f^{n-1}\right) \in \left\langle I, tf-1 \right\rangle_{k\left[\textbf{x},t\right]}. \end{align}
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