show that the ideal is prime 
Let $K[a,b,c,d]$ be a polynomial ring over a field and $A=(ad-bc,a+d)$ an ideal of $K[a,b,c,d]$. Show that $A$ is prime. 

What are the basic methods of showing that an ideal is prime? 
 A: We have $$K[a,b,c,d]/(ad-bc,a+d)=\frac {K[a,b,c,d]/(a+d)}{(\bar a \bar d-\bar b\bar c)}= \frac {K[\bar a,\bar b,\bar c]}{(-\bar a ^2 -\bar b\bar c)}=\frac {K[\bar a,\bar b,\bar c]}{(\bar a ^2 +\bar b\bar c)}$$ since $\bar d=-\bar a$ in $K[\bar a,\bar b,\bar c]:=K[a,b,c,d]/(a+d)$.
In the polynomial ring $K[\bar a,\bar b,\bar c]$ the polynomial $\bar a ^2 +\bar b\bar c$ is irreducible, hence generates a principal prime ideal and thus $\frac {K[\bar a,\bar b,\bar c]}{(\bar a ^2 +\bar b\bar c)}=K[a,b,c,d]/(ad-bc,a+d)$ is a domain, which proves that $A=(ad-bc,a+d)$ is prime.
(I have used that in a UFD, here $K[\bar a,\bar b,\bar c]$,  any irreducible element generates a prime ideal).   
Edit
At the request of DonAntonio in the comments here is why $\bar a ^2 +\bar b\bar c$ is irreducible in $\frac {K[\bar a,\bar b,\bar c]}{(\bar a ^2 +\bar b\bar c)}$:
Given any ring  $S$ and its polynomial ring $S[T]$, the polynomial $T^2-s\in S[T]$ is irreducible  if and only if $s$ is not a square in the ring $S$.
Just apply this to the case $S=K[\bar b,\bar c], T=\bar a$ and $s=\bar b\cdot \bar c$.   
A: Observe that in the quotient ring $\;R/A\;,\;\;R=K[a,b,c,d]\;$, we have that $\;ad=bc\;,\;\;a=-d\;$ , so we can in fact write $\;bc=-a^2\implies b=-\frac{a^2}b\;$ in the quotient, and then try to make a first characterization of the elements in the quotient, which can be seen as polynomials $\;g(x,y,z,w)\;$ , such that $\;w=-x,\,x^2=-bc,\,c=-\frac{a^2}b\;$
For example, we have that the polynomial $\;a+b+c+d\in R\;$ is mapped under the canonical projection to the element $\;x+y-\frac{x^2}y-x=y-\frac{x^2}y\;$. 
Thus, we can already see the quotient is isomorphic with a ring of polynomials in two variables $\;x,y\;$ under the above relations , and thus $\;R/A\;$ is an integer domain $\;\iff A\;$ is a prime ideal.
