Is the ideal $I=(x_1 x_5 - x_2 x_4 , x_1 x_6 - x_3 x_4)$ of $k[x_1,...,x_6]$ a radical ideal? Is it a prime ideal? Is the ideal $I=(x_1 x_5 - x_2 x_4 , x_1 x_6 - x_3 x_4)$ of $k[x_1,...,x_6]$ a radical ideal? Is it a prime ideal?  
thanks
 A: It is not prime since
$$x_1(x_3x_5-x_2x_6)=x_3(x_1x_5-x_2x_4)-x_2(x_1x_6-x_3x_4)\in I$$
but $x_1\notin I$ because every nonzero element of $I$ has degree at least $2$, and $x_3x_5-x_2x_6\notin I$ because every element of degree $2$ in $I$ is a $k$-linear combination of $x_1x_5-x_2x_4$ and $x_1x_6-x_3x_4$ and so cannot involve either of the terms $x_3x_5$ or $x_2x_6$. 
Note that this is taking advantage of the fact that $I$ is homogeneous.
A: If you compute a Gröbner basis for $I$, you will get that you need one additional generator, namely $x_1x_3x_5-x_1x_2x_6=x_1(x_3x_5-x_2x_6)$. Thus $x_1$ vanishes on the zero set of the ideal, but $x_1 \not \in I$, as in Alex Becker's anser above. 
Your ideal in fact decomposes as $I= I_{minors} \cap \langle x_1, x_4 \rangle$, where $I_{minors}$ is the ideal of $2 \times 2$-minors of a generic matrix
$$ \begin{bmatrix}x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6\end{bmatrix}.$$
So the "reason" the ideal is not prime is that you didn't include all minors.
ADDED To see the decomposition $I=I_{minors} \cap \langle x_1,x_4 \rangle$, we show first the $\subseteq$-direction. But this is clear, because the generators of $I$ are a subset of the generators of $I_{minors}$, and they clearly lie in $\langle x_1,x_4\rangle$. For the other direction, one notices that already two of the generators of $I_{minors}$ lie in $I$. The last one, $x_2x_6-x_3x_5$, can be multiplied by elements of $\langle x_1,x_4\rangle$ to give $x_1x_3x_5-x_1x_2x_6$, wich lies in $I$, because it is part of a Gröbner basis of $I$, as explained above. The the decomposition is shown.
Now it is a result that any determinantal ideal is prime (try Google), and so $I$ is an intersection of prime ideals. Thus $I$ is radical, because an ideal is radical if and only if it decomposes as an intersection of radical ideals (easy exercise!). (for a hint: suppose $f^k \in I$. Then $f^k$ is also contained in all the components of $I$, but they are all radical...)
