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I'm trying to prove this ideal

$$I=(x^2+y^2+x,x+y+xy)\subset \mathbb C[x,y]$$

is prime.

I supposed that $I$ is prime and I'm using the classical method to prove $I$ is prime: If $ab\in I$, then $a\in I$ or $b\in I$. I didn't have any success with this method and I would like to know which are the tools and strategies I could use to solve this question.

I really need help.

Thanks in advance.

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If you know some algebraic geometry:

If $I$ is a prime ideal then the variety it cuts out $V(I)$ is an irreducible variety.

Points in the variety satisfy the equations $x+y+xy=0$ and $x^2+y^2+x=0.$ If $x=0$ then $y=0$ and if $y=0$ then $x=0.$

So now suppose we have a point $(x,y)$ on the variety with both $x$ and $y$ non-zero. Then $x+y+xy=0 \implies y = -x/(1+x),$ which put into the other equation yields $x(x^3+3x^2+4x+1)=0.$ We can do a similar thing to get an equation that $y$ must satisfy.

The conclusion is: $V$ has finitely many possible $x$ and $y$ values, and hence is a finite set. It also has more than one point: $(0,0)$ and $(a, -a/(1+a) )$ (where $a$ is a root of $x^3+3x^2+4x+1=0$) are points in $V.$ So $V$ is not irreducible, and $I$ is not a prime ideal.

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A remark : It is a theorem of Bézout that if you have two plane projective curves given by polynomials in two variables, the number of solutions is the product of the degrees (counting multiplicities and points at infinity). So in this case you expect at most four points in the algebraic set associated to your ideal $I$.

A quick way to check that $I$ is not prime is just to notice that the algebraic set associated to $I$ contains more than one point (by the remark above, it contains finitely many points), because containing more than one point means the algebraic set is reducible, hence its ideal cannot be prime. I convinced myself using Wolfram Alpha that there are four solutions in this case. This is not really a full proof since it uses the Bézout sledgehammer and there are more elementary techniques to do this (see other answers), but it's always cool to keep it in mind.

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It's not hard to prove that two plane curves (which don't share a common component) have only finitely many points in common without Bezout. See the first half of my answer here. So the idea in your answer is a good one to keep in mind even for someone who doesn't know Bezout. – Ragib Zaman Jun 10 '14 at 8:40
@RagibZaman : That's a nice trick, I like it! – Patrick Da Silva Jun 10 '14 at 14:57

Suppose that $I$ is prime. Obviously, $I\subseteq (x,y)$.

If $I=(x,y)$, then $x\in (x^2+y^2+x,x+y+xy)$, and therefore $$x=(x^2+y^2+x)g(x,y)+(x+y+xy)h(x,y).$$ But this is not possible: replace $(x,y)$ by $(a,b)$ where $a\ne 0$, $a^2+b^2+a=0$, $a+b+ab=0$, and get a contradiction.

If $I\subsetneq(x,y)$, then $I$ contains a principal prime ideal. (Note that $K[x,y]$ is factorial.) But in $K[x,y]$ there are no chains of prime ideals of length $3$, so $I$ is principal. It's easily seen that this is not possible since both generators of $I$ are irreducible polynomials.

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A Groebner basis for $I$ is $\{-y+y^2+2 y^3+y^4, x+y^2+y^3\}$ The first factors as $y(-1+y+2y^2+y^3)$. A Groebner basis of $I \cup \{y\}$ is $\{x, y\}$ and a Groebner basis of $I \cup \{-1+y+2y^2+y^3\}$ is $\{1+4x+3x^2+x^3, -1-2x-x^2+y\}$. Since neither of these is the Groebner basis for $I$, even though all of these bases are computed under the same monomial order, we find $y(-1+y+2y^2+y^3) \in I$, but also $y \not \in I$ and $-1+y+2y^2+y^3 \not \in I$. $I$ is therefore not a prime ideal.

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Interesting. Do you know a good source for Groebner bases for someone with a good background in algebra looking to get into algebraic geometry? – drawnonward Jun 10 '14 at 7:06
@drawnonward Cox, Little, O'Shea "Ideals, Varieties and Algorithms." – Casteels Jun 10 '14 at 13:02

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