Let $k$ be a field. Consider the ideals $I_1=(x),I_2=(y),J=(x^2,y)$ of $R=k[x,y]/(xy,y^2)$. Show that the homogeneous elements of $J$ are contained in $I_1\cup I_2$ but that $J\not\subset I_1$ and $J\not\subset I_2$.

First of all, what is meant by "homogeneous elements of $J$"? The notion of homogeneous element makes sense in a graded module $M=\oplus_{i=-\infty}^{\infty}M_i$ over a graded ring $R=R_0\oplus R_1\oplus\dots$. The ideal $J$ is an $R$-module, but how is the grading of $J$ and $R$ defined?

Also, whichever the definition of a homogeneous element is, I think the element $x+y$ must be homogeneous according to the definition. But it doesn't lie in the union $I_1\cup I_2$...


The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups $$A_n:=\bigoplus_{i=0}^nx^iy^{n-i}k,$$ give the grading on $k[x,y]=\bigoplus_{n\geq0}A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that $$B_0=k,\qquad B_1=\overline{x}k\oplus\overline{y}k \qquad\text{ and }\qquad B_n=\overline{x}^nk \text{ for }n>1.$$ Then $R=\bigoplus_{n\geq0}B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.

As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.

  • $\begingroup$ I found out I don't quite understand the grading of $k[x,y]$. By your definition, $A_1=kx\oplus ky$, $A_2=kx^2\oplus kxy\oplus y^2$. They are both abelian groups But the definition of the graded ring says that one must have $R_iR_j\subset R_{i+j}$. How is multiplication of elements of $A_1$ by elements of $A_2$ defined? The elements of $A_1$ are of the form $(a_1x,a_2y)$, and those of $A_2$ are of the form $(b_1x^2,b_2xy,b_3y^2)$; how to multiply them? Also, how does the grading of $R$ give a grading on its submodule? $\endgroup$ – user419669 Mar 12 '19 at 21:21
  • $\begingroup$ These groups are still subsets of their respective rings; their elements are multiplied as elements of the ring. So writing more conventionally, two elements of $A_1$ and $A_2$ are multiplied as \begin{eqnarray*} (a_1x,a_2y)\cdot(b_1x^2,b_2xy,b_3y^2)&=&(a_1x+a_2y)(b_1x^2+b_2xy+b-3y^2)\\ &=&a_1b_1x^3+(a_1b_2+a_2b_1)x^2y+(a_1b_3+a_2b_2)xy^2+a_2b_3y^3\\ &=&(a_1b_1,a_1b_2+a_2b_1,a_1b_3+a_2b_2,a_2b_3)\in A_3. \end{eqnarray*} $\endgroup$ – Servaes Mar 12 '19 at 23:34
  • $\begingroup$ For some more context: The grading on $k[x,y]$ is by degree; an element of $k[x,y]$ is homogeneous of degree $n$ if it is a linear combinations of monomials of degree $n$. For example the polynomials $$x^3+3xy^2+4y^3\qquad\text{ and }\qquad x^7+5x^3y^4+2xy^6,$$ are homogeneous of degree $3$ and $7$, respectively. But the polynomial $x^2+3y$ is not homogeneous at all, because the individual monomials do not have the same degree. The subgroup $A_n$ is then the subset of all elements of $k[x,y]$ of degree $n$, including $0$. This makes clear that $A_iA_j\subset A_{ij}$. $\endgroup$ – Servaes Mar 12 '19 at 23:38
  • $\begingroup$ The second description is how I imagined grading on $k[x,y]$. The notation with direct sums (like $kx\oplus kx$) is confusing because formally speaking, this is not a subgroup of the ring $k[x,y]$ (because the ring does not contain any ordered pairs, it contains elements of different nature). Is there a way to formalize the statement that $A_i$ is a subgroup of $k[x,y]$? And how does the grading on $R$ give a grading on $J$? Can I think of homogeneous elements on $J$ as homogeneous polynomials in the residue classes of $x,y$? $\endgroup$ – user419669 Mar 12 '19 at 23:59
  • $\begingroup$ The direct sums are subgroups of $k[x,y]$, in fact they are even subspaces ($k$-vector spaces). If you distinguish between internal and external direct sums, perhaps read this. It helps not to think of a direct sum as a product of sets with an operation. But if you prefer, you could write $A_1=xk+yk$ though the notation gets uglier if you can't write $A_n=\bigoplus_{i=0}^nx^iy^{n-i}k$. $\endgroup$ – Servaes Mar 13 '19 at 0:06

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