I want to show that if $I=(x_1-x_4, x_1^2- x_4 x_2, x_1^2 - x_4 x_3)$ then $I+(x_1-1) = (x_1-1, x_2-1, x_3-1, x_4-1 )$ for $K[x_1,\dots,x_4]$ with $K$ a field.
I tried showing that $x_i -1$ can be written as sum of someone in $I$ and some multiple of $x_1-1$ $\forall i$. I only can do this for $i=1,4$. I do not know if I am aproaching this problem as I should.
Hint $ $ use $\ (a,b,c\ldots) = (a,\, b\bmod a,\, c\bmod a\ldots)\,$ as in the Euclidean algorithm, so
$$\begin{align} J &= \ (\color{}{x_1}-1,\ \color{#c00}{x_1}-\,x_4,\, \color{#c00}{x_1}^2- x_2 x_4,\, \color{#c00}{x_1}^2 - x_3 x_4)\\ &=\, (x_1-1,\ \, \color{#c00}1\: -\ x_4,\ \ \color{#c00}1\ -\,\ x_2\color{#0a0}{x_4},\, \ \color{#c00}1\ -\ \,x_3 \color{#0a0}{x_4})\ \ \ {\rm by}\ \ \color{#c00}{x_1\equiv 1}\!\!\!\pmod{\!J}\\ &= \,(x_1-1,\, \ 1\ -\ x_4,\ \ 1\ -\ \ x_2,\ \ \ \ \ \, 1\ -\ \, x_3)\ \ \ \ \ \ \ {\rm by}\ \ \color{#0a0}{x_4\equiv 1}\!\!\!\pmod{\!J} \end{align}\qquad\qquad$$