I am trying to understand how $I_n$-fibres appear in an elliptic surface by performing a sequence of blow-ups. To be concrete, I am looking at the following elliptic surface given in Weierstrass normal form: $$y^2=x^3+x^2+t^3x+t^4$$
This surface has an $I_4$ fibre when $t=0$, according to Tate's algorithm, and I am trying to see this concretely. I feel like I understand how to do the first blow up:
I cover the blow-up with two charts, the first of which is given by $$t=t,y=y_1t, x=x_1t$$
Substituting these into the equation above, I find that $$t^2\left[y_1^2-tx_1^3-x_1^2-tx_1-t^2\right]=0$$
Thus, the strict transform of my surface is given by the equation in brackets, which is still singular. The $t=0$ fibre is the pair of lines $y_1=\pm x_1$ in this chart.
This is what happens in another chart, if I use coordinates $$t=x''t'',y=y''x'', x=x''$$ The strict transform works out to $$y''^2=x''+1+x''^2t^3+t''^4x''^2$$
Analyzing the $t=t''x''=0$ fibre in this chart, we have two pieces: $$t''=0, y''^2=x''+1$$ $$x''=0,\ y''^2=1$$
If I check how these all glue, then the pair of lines that I found in the first chart becomes the pair of lines here, and a similar thing happens with the last chart.
In summary, I am convinced that at this point, the $t=0$-fibre on my current surface is three rational curves, arranged in a triangle.
At this point, I know what I need to do, but I don't think I'm doing it right. The surface I just obtained is still singular at the point $(x_1,y_1,t)=(0,0,0)$, so I need to blow it up. My intuition is that this will "separate" the two lines by introducing one more rational component meeting them each once. In total, this leaves 4 rational components arranged in a polygon, which is exactly what I should be getting. However, that is not what happens:
I perform another blow-up by looking at the following chart: $$t=t, x_1=x_2t, y_1=y_2t$$ I find that the strict transform is $$y_2^2=t^2x_2^3+x_2^2+x_2t+1$$
This is now smooth, and the $t=0$ fibre here is $$y_2^2=x_2^2+1,$$ which is a single rational component, as I had hoped. Now, the part that's not working is seeing how this component intersects the other three components that I found earlier.
The pair of lines was given, in $(x_1,y_1,t)$ coordinates by $t=0,y_1=\pm x_1$. If I consider these equations in $(x_2,y_2,t)$ coordinates, I get $t=0, y_2=\pm x_2$. But neither of these lines intersect the rational component I found, so there's no way I can build an $I_4$ fibre from this!
If anyone can tell me what I'm doing wrong/ what I'm not seeing about this picture, I would be very appreciative.