using smoothness to deduce reducedness Let $F_1,\dots,F_s$ be homogeneous polynomials in the ring $S=k[X_0,\dots,X_n], s < n$ and suppose that they form a regular sequence. Suppose also that $X = V(F_1) \cap \dots \cap V(F_s)$ is smooth. How can the smoothness of $X$ be used to deduce that $(F_1,\dots,F_n)$ is radical or equivalently that $k[X_0,\dots,X_n]/(F_1,\dots,F_s)$ is reduced? In classic terms, smoothness of $X$ is equivalent to $O_{X,P}$ being regular for every point $P \in X$. The issue is that $O_{X,P}$ is related to the homogeneous coordinate ring of $X$ rather than $k[X_0,\dots,X_n]/(F_1,\dots,F_s)$. 
PS: This is the missing step to my understanding of the ideal generated by general polynomials is radical.
 A: Manos, I know the frustration, so let me make another attempt in a different direction. I will do a simple case, where $n=3,s=2$. So, we have $V(F_1,F_2)=X\subset\mathbb{P}^3$, a smooth, irreducible complete intersection curve. Let $I\subset\mathcal{O}_{\mathbb{P}^3}$ be its ideal sheaf. Then, we have the Koszul resolution, $0\to \mathcal{O}_{\mathbb{P}^3}(-\deg F_1-\deg F_2)\to \mathcal{O}_{\mathbb{P}^3}(-\deg F_1)\oplus \mathcal{O}_{\mathbb{P}^3}(-deg F_2)\to I\to 0$, which by using the the fact that $H^i(\mathcal{O}_{\mathbb{P}^3}(l))=0$ for $i=1,2$ and all $l$ gives $H^1(I(l))=0$ for all $l$ and the map $H^0(\mathcal{O}_{\mathbb{P}^3}(-\deg F_1+l)\oplus \mathcal{O}_{\mathbb{P}^3}(-\deg F_2+l))\to I(l)$ is onto for all $l$. Thus, we see that $\oplus I(l)\subset\oplus\mathcal{O}_{\mathbb{P}^3}(l)=k[x_0,x_1,x_2,x_3]$ is just the ideal $(F_1,F_2)$. From the long exact sequence coming from $0\to I\to\mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_X\to 0$, using the vanishing of $H^1$, we get an exact sequence, $0\to (F_1,F_2)\to k[x_0,x_1,x_2,x_3]\to \oplus H^0(\mathcal{O}_X(l))\to 0$. Using the fact that $X$ is integral, easy to check that so is $\oplus H^0(\mathcal{O}_X(l))$, but this is precisely $k[x_0,x_1,x_2,x_3]/(F_1,F_2)$. 
A: First of all we can take $s \le n$. Since $F_1,\dots,F_s$ is a regular sequence, we have that $\operatorname{height}(I)=s$, where $I=(F_1,\dots,F_s)$, and also $R=S/I$ is Cohen-Macaulay.
Let $J$ be the Jacobian ideal of $F_1,\dots,F_s$, i.e. the ideal generated by all $s \times s$ minors of the Jacobian matrix of the $F_i$. If $X = Z(F_1,\dots,F_s)$ is smooth, then $J \not\subset \mathfrak{m}_x, \, \forall x \in X$ and so in the quotient ring $R$ we must have $\operatorname{height}(\bar{J}) \ge 1$. Finally, according to Theorem 18.15a in Eisenbud (CA, 2004) $I$ is reduced if and only if $\operatorname{height}(\bar{J}) \ge 1$. 
A: If $F_1,\ldots,F_s$ are general, by Bertini theorem, we get that $\dim X=n-s$ and it is smooth. Thus, $k[x_0,\ldots,x_n]/(F_1,\ldots,F_s)$ is a complete intersection (follows from $\dim X=n-s$) and outside the irrelevant  maximal ideal, this ring is regular. Since complete intersection implies the ring is Cohen-Macaulay, to see whether it is reduced, it suffices to do so at the minimal primes and this follows from the regularity at all the relevant primes.
