How do I show that the three polynomials $f_1 = t^2 + x^2 - 2$, $f_2 = tx - 1$, $f_3 = t^3 + 5tx^3 + 1$ generate the unit ideal in $\mathbb{C}[t, x]$?
Artin mentions two approaches: by showing that they have no common zeros, and also by writing $1$ by a linear combination of $f_1$, $f_2$, $f_3$, with polynomial coefficients.
Algebraic Geometry Approach
We claim $f_1,f_2,f_3$ have no common $($complex$)$ zeros. Suppose otherwise, that there exists a common zero $(u,v)$; then $$0 = f_1 - 2f_2 = (u^2+v^2-2) - 2(uv - 1) = (u-v)^2,$$ so $u-v=0$. Now we have $0 = f_2 = u^2-1$, so $u = \pm1$. But $$0 = f_3 = u^3+5uv^2+1 = 6u^3 + 1 = \pm 6 + 1 \ne0,$$ contradiction.
By Corollary 11.9.4 in Artin, $$(f_1,f_2,f_3) = (1) = \mathbb{C}[x,y].$$
Direct Approach
We have $$f_3 = t^3+5tx^2+1 = t(f_1-x^2+2)+5tx^2+1 = tf_1 + 2t+1+4tx^2 $$$$= tf_1 + 2t+1+4x(f_2+1) = [tf_1+4xf_2] + [2t+4x+1].$$
Now, $$-tf_1 -4xf_2 + f_3 = 3(t+x) - (t-x) + 1.$$ Let $$p(t,x) = 3(t+x) - (t-x) + 1,\text{ }q(t,x) = p(x,t)p(-t,-x)p(-x,-t);$$then $$tq(t,x) f_1 - 4xq(t,x) f_2 + q(t,x) f_3 $$$$= p(t,x)q(t,x)$$$$= ([3(t+x)+1]^2 - (t-x)^2)([-3(t+x)+1]^2 - (t-x)^2)$$$$ = [9(t+x)^2+1-(t-x)^2]^2 - 6^2(t+x)^2 $$$$ = [8(f_1+2)+20(f_2+1) + 1]^2 - 36[(f_1+2)+2(f_2+1)] $$$$ = [8f_1+20f_2+37]^2 - 36[f_1+2f_2+4] $$$$ = (37^2-4\cdot36) + 74(8f_1+20f_2) + (8f_1+20f_2)^2 - 36[f_1+2f_2] $$$$ = 35^2 + f_1[556 + 64f_1 + 160f_2] + f_2[1408 + 160f_1 + 400f_2],$$ so $\alpha f_1 + \beta f_2 + \gamma f_3 = 1$, where $$\alpha = -\frac{1}{35^2}(tq(t,x) + 556 + 64f_1 + 160f_2),$$$$\beta = -\frac{1}{35^2}(4xq(t,x) + 1408 + 160f_1 + 400f_2),$$$$\gamma = \frac{1}{35^2}q(t,x).$$