Proof that ideals are irreducible

Question

In Hassett's book Introduction to Algebraic Geometry he states:

$$ (x^2, xy, y^2 ) = (y+x, x^2) \cap (x,(y+x)^2)$$ and both ideals are irreducible (i.e., not writable as intersections of non-trivial ideals). I was trying to prove it, but I have no idea. It is easy to prove $\subset$ just by showing the generators are inside the right-hand site. But what about the other direction? Is there any possibility to write down the generators of an intersection of ideals?

And how can I prove that these ideals are indeed irreducible?

Answer 1

Given $f \in (x+y, x^2) \cap (x,(x+y)^2)$, then $$ f = a(x+y) + b x^2 = cx + d(x+y)^2 $$ for some $a,b,c,d \in k[x,y]$. Grouping terms, we have \begin{align*} x(bx-c) = bx^2 - cx = d(x+y)^2 - a(x+y) = (x+y)(d(x+y) - a) \, . \end{align*} Since $x$ is prime and doesn't divide $x+y$, then $x \mid (d(x+y) - a)$, so $$ d(x+y) - a = hx $$ for some $h \in k[x,y]$. Then $a = -hx + d(x+y)$, so \begin{align*} f &= a(x+y) + bx^2 = (-hx + d(x+y))(x+y)\\ &= -hx^2 - hxy + d(x^2 + 2xy + y^2) + bx^2 \in (x^2, xy, y^2) \, . \end{align*}

As a note, there is an algorithmic way to compute the intersection of ideals using Gröbner bases. This is given in Proposition $30$ of $\S9.6$ of Dummit and Foote:

Proposition 30. If $I$ and $J$ are any two ideals in $F[x_1, \ldots, x_n]$ then $tI + (1 - t)J$ is an ideal in $F[t, x_1, \ldots, x_n]$ and $I \cap J = (tI + (1 - t)J) \cap F[x_1, \ldots, x_n]$. In particular, $I \cap J$ is the first elimination ideal of $t I + (1 - t) J$ with respect to the ordering $t > x_1 > \cdots > x_n$.

Computing as follows using SageMath (here is a link to a SageMathCell)

R.<t,x,y> = PolynomialRing(QQ,3,order="lex")
I = ideal([x+y, x^2])
J = ideal([x, (x+y)^2])
IJt = t*I + (1-t)*J
IJt.groebner_basis()

we find that $\{t x - x, t y + x, x^{2}, x y, y^{2}\}$ is a Gröbner basis for $tI + (1 - t)J$, hence $$ I \cap J = (tI + (1 - t)J) \cap k[x,y] = (x^{2}, x y, y^{2}) $$ by the Proposition, as we found before.

This is just a rephrasing of @imtrying46's excellent answer, but one that I found helpful. Recall that the Lattice Isomorphism Theorem gives an inclusion-preserving bijection between the ideals of $k[x,y]$ containing $(x+y, x^2)$ and the ideals of the quotient $k[x,y]/(x+y, x^2)$. Note that $$ \frac{k[x,y]}{(x+y, x^2)} \cong \frac{k[x]}{(x^2)} $$ by the map sending $y \mapsto -x$. The only proper, nonzero ideal of $k[x]/(x^2)$ is $(x)$, hence we conclude that $$ (x) + (x+y, x^2) = (x, x+y, x^2) = (x,y) $$ is the only proper ideal of $k[x,y]$ strictly containing $(x+y, x^2)$, just as in @imtrying46's answer.

Answer 2

Intersections of ideals can be quite tricky to compute, but it can be done by hand. Here, note the following: $(x^2,xy,y^2)$ is exactly the set of polynomials where the coefficients of $1$, $x$ and $y$ are equal to $0$. In other words, it is the set of polynomials $p$ such that $p(0,0)=0$, $\partial_x p(0,0)=0$ and $\partial_yp(0,0)=0$.

Now let $p\in (y+x,x^2)\cap (x,(y+x)^2)$ be arbitrary and write $p=a(y+x)+bx^2=cx+d(y+x)^2$ for polynomials $a,b,c,d\in k[x,y]$. It is immediately clear that $p(0,0)=0$. Now for any polynomial $q$, denote $\partial_x q=q_x$ and $\partial_y q=q_y$. Then by the Leibnitz rule we have $$ p_x=a_x(y+x)+a+b_xx^2+2bx=c_xx+c+d_x(y+x)^2+2d(y+x) $$ and $$ p_y=a_y(y+x)+a+b_yx^2=c_yx+d_y(y+x)^2+2d(y+x). $$ So we already obtain $p_y(0,0)=a(0,0)=0$ by the last string of equations. Then, it follows that $p_x(0,0)=a(0,0)=0$ by the first equation. So $p\in(x^2,xy,y^2)$, proving the reverse inclusion.

Now to see that e.g. $(y+x,x^2)$ is irreducible, you can proceed as follows: any class $p+(y+x,x^2)$ has a unique element $q$ of the form $a+bx$ with $a,b\in k$. Indeed, as $x+y$ divides $p(x,y)-p(x,-x)$, we may replace $p$ by the polynomial $p(x,-x)$ depending only on $x$, and then with $x^2$ we can kill everything but $a+bx$. So if $I\subsetneq k[x,y]$ is a proper ideal strictly containing $(y+x,x^2)$, it must contain some element of the form $a+bx\neq 0$. As it is proper, we must have $b\neq 0$, and hence up to multiplying by a unit, we may assume $b=1$. So $I$ contains $a+x$ and $x^2$, hence also $a^2=x^2-(x+a)(x-a)$, and thus $a=0$. Therefore $I$ contains $x$ and $(y+x,x^2)$, and thus also $(x,y)=(x)+(y+x,x^2)$. As $(x,y)$ is maximal and $I$ is proper, it follows that $I=(x,y)$.

In conclusion, the only proper ideal strictly containing $(y+x,x^2)$ is $(x,y)$. In particular, $(y+x,x^2)$ is irreducible. Now under the automorphism of $k[x,y]$ sending $x\mapsto x+y$ and $y\mapsto -y$, the ideal $(y+x,x^2)$ is mapped to the ideal $(x,(y+x)^2)$, hence this is irreducible too.