Prime ideal after modding out by homogenizing variable implies original ideal prime?

Question

Let $R:=\mathbb{C}[x_1,\ldots,x_d]$ be a polynomial ring. Let $I=(f_1,\ldots,f_n)$ be an $R$-ideal. Let $f^*_i$ be the homogenization of $f_i$ in $R[t]$.

Question: If $(\overline{f_1^*},\ldots,\overline{f_n^*})$ is prime, where $\overline{f_i^*}$ denotes the image of $f_i^*$ in $R$ under the map $t\mapsto 0$, is $I$ prime?

Edit: I changed the question to be specifically about homogenizing. I also changed notation to use $t$ to hopefully clarify the homogenization.

I want to rephrase the question using the filtration language:

First we define an orderly $\mathbb{Z}$-filtration $\mathcal{F}$. This means $\mathcal{F}$ satisfies the four filtration axioms:

1. $F_g\subset F_h$ for $g<h$
2. $\bigcup_{g\in\mathbb{Z}} F_g R = R$
3. $(F_g R)(F_h R)\subseteq F_{g+h}R$ for all $g,h\in\mathbb{Z}$
4. $1\in F_0R\setminus F_{<0}R$ (this is set minus),

where $F_{<g}:=\bigcup_{h<g}F_hR$,

along with the orderly axiom:

5. For all nonzero $a\in R$ there exists a smallest element $g\in\mathbb{Z}$ such that $a\in F_g R$. The smallest element $g$ is then called the order of $a$. Note that an orderly $\mathbb{Z}$-filtration is separated.

Now we define a degree filtration by letting $\alpha=(\alpha_1,\ldots,\alpha_d)$ be a vector of positive integers and assign $R$ the structure of a positively graded $\mathbb{C}$-algebra by setting $\deg_\alpha x_i=\alpha_i$ for $1\leq i\leq d$. Define $\mathcal{F}^\alpha=\{F_i^\alpha\}$ by setting $$F_i^\alpha R=\bigoplus_{j\leq i}R_j.$$ Here $R_j$ denotes the $j^{th}$ homogeneous component with respect to the $\alpha$-graduation.

Define the associated graded ring $$\text{gr}_{\mathcal{F}^a}(R)=\bigoplus_{g\in\mathbb{Z}}F_g R/F_{<g}R.$$ Define the initial form of $a\in R$ of order $g$ to be the residue class $\text{in}_{\mathcal{F}^\alpha}(a)=a+F_{<g}R$.

The question now becomes: If $(\text{in}_{\mathcal{F}^\alpha}(f_1),\ldots,\text{in}_{\mathcal{F}^\alpha}(f_n))$ is prime, then is $(f_1,\ldots,f_n)$ prime?

Answer 1

Geometry hints at a counterexample: the union of the two lines $y = x$ and $y = 1$ is not irreducible, but its intersection with the $x$-axis is a single point. So, translating into commutative algebra, consider the ideal $\langle (y - x)(y - 1) \rangle \subset \mathbb{C}[x, y]$, which is not prime. Its image under the projection map is the prime ideal $\langle x \rangle \subset \mathbb{C}[x]$.