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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?

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    One basic method would be to find a ring homomorphism $K[a,b,c,d] \rightarrow S$ into some integral domain $S$ and with kernel $A$. Not sure, if this is the way to go, though. – 2017-02-21
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    Note that $ad-bc=\det\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and $a+d = \operatorname{tr}\begin{pmatrix} a & b \\ c & d\end{pmatrix}$. Not sure how this is helpful, though. – 2017-02-21
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    @lisyarus: I also had this thought but there is no relation between the polynomial ring and the ring generated by the matrix, the multiplication rules conflict. But maybe there is a "trick"? – 2017-02-21
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    @MarcBogaerts Yes, conflicting multiplication confuses me, too. But there should be some trick! I hate coincidences. – 2017-02-21

2 Answers 2

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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 $s=\bar b\cdot \bar c$.

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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.

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    $bc=-a^2\implies b=-\frac{a^2}b$ you mean $bc=-a^2\implies b=-\frac{a^2}c$? why can you divide by $c$? – 2017-02-21
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    @D The **symbol** $\;-\frac{a^2}c\;$ is just that: *a symbol*. Put $\;y:=-\frac{a^2}c\;$ instead if you want. Still, and unless I am wrong, you get polynomials in two variables. That's all that matters. – 2017-02-21
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    Dear DonAntonio, the quotient ring $R/A$ is not isomorphic to any ring of polynomials in two variables – 2017-02-21
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    @GeorgesElencwajg Perhaps not, George. Thanks...yet I still would want to see the proof of this. What you show in your answer is the unfounded claim that $\;\overline a^2+\overline b\overline c\;$ is irreducible in that quotient ring, which would be nice also to see its proof. For what is worth, I don't really think you're wrong and I am right. I'm only asking for *a proof* of your claims. Thanks, again. – 2017-02-21
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    $K[X,Y,Z]/(X^2-YZ)$ is not a UFD, so it can't be isomorphic to a ring of polynomials over a field. – 2017-02-21
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    @user26857 Thanks, that's an excellent point that I oversaw as I didn't go for a *partial* quotient ring. I shall leave this answer and in particular the comment for others to read. – 2017-02-21
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    Dear DonAntonio, at your request I have explained in an Edit why $\bar a ^2 +\bar b\bar c$ is irreducible. By the way, a geometric reason why $K[X,Y,Z]/(X^2-YZ)$ is not a polynomial ring is that its associated scheme is smooth while the associated scheme to a polynomial ring is affine spavce, which is smooth. Of course @user26857's explanation is much more elementary and thus much better! – 2017-02-21