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In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions. See for example Proposition 1.1.2 in Winfried Bruns and Jurgen Herzog's book Cohen-Macaulay Rings. The argument is basically that an element being regular means multiplication by it is injective, and flat base change preserves injectivity, followed by induction on the length of the sequence.

In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $\dim V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions. See for example Proposition 1.1.2 in Winfried Bruns and Jurgen Herzog's book Cohen-Macaulay Rings. The argument is basically that an element being regular means multiplication by it is injective, and flat base change preserves injectivity, followed by induction on the length of the sequence.

In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions.

In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions. See for example Proposition 1.1.2 in Winfried Bruns and Jurgen Herzog's book Cohen-Macaulay Rings. The argument is basically that an element being regular means multiplication by it is injective, and flat base change preserves injectivity, followed by induction on the length of the sequence.

In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions.

In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions.