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I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

If $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you generically reduced assumption, then $Y$ is reduced. (This is just restating what you are saying in #3).

A simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

If $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you generically reduced assumption, then $Y$ is reduced. (This is just restating what you are saying in #3).

A simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$, where $Y$ is a complete intersection in $X$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

So, if $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you genericly reduced assumptiom, then $Y$ is reduced. (This is just restating what you are saying in #3).

So, a simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see this below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

If $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you generically reduced assumption, then $Y$ is reduced. (This is just restating what you are saying in #3).

A simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

So, if $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you genericly reduced assumptiom, then $Y$ is reduced. (This is just restating what you are saying in #3).

So, a simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X = \mathrm{Spec} k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X$ is smooth away from that line, so it is $R_1$. We will see this below that it is $S_2$ and hence normal.

Next let $Z = Z(w)\subset X$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

So,

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

So, if $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you genericly reduced assumptiom, then $Y$ is reduced. (This is just restating what you are saying in #3).

So, a simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see this below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.