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Let $R$ be a ring. Let $I = \langle x + y \rangle$, $J = \langle x - y \rangle$ be ideals of $R[x,y]$.

What's $I + J$ in this case? By definition $I + J = \{ i + j \mid i \in I, j \in J \}$. My first thought was that $I + J = \langle 2x \rangle$ but since $x(x+y) \in I$ and $y(x-y) \in J$ and $x(x+y) + y(x-y) = (x-y)^2 \notin \langle 2x \rangle$ this must be wrong. Is there a formula for $I + J$ if $I,J$ are generated by some set of elements?

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As $I+J$ is the smallest ideal containing $I$ and $J$ you have $I + J = \langle M \cup N \rangle$ for $I = \langle M\rangle$ and $J = \langle N \rangle$, in your example $I + J = \langle x+y, x-y\rangle$. – martini Apr 11 '12 at 12:54
@martini Thank you. This answers my question. – Rudy the Reindeer Apr 11 '12 at 12:56
So I will post it as an answer then. – martini Apr 11 '12 at 12:58
You might want to give it a better title, for two reasons: (1) It is a bad habit to use LaTeX only titles; and more importantly (2) Just looking at the title tells me absolutely nothing about the content of the question. – Asaf Karagila Apr 11 '12 at 13:02
@AsafKaragila Is this better? – Rudy the Reindeer Apr 11 '12 at 13:17
up vote 4 down vote accepted

We know, that $I+J$ is the smallest ideal containing $I$ and $J$. If we are given $I$ and $J$ by generators, say $I = \langle M \rangle$ and $J = \langle N \rangle$, we have that $I+J$ is the ideal generated by all these generators, i. e. we have $I + J = \langle M \cup N \rangle$.

In your example above we have therefore $I + J = \langle x+y, x-y\rangle$.

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As explained in martini's answer, $I+J=\langle x+y, x-y\rangle$.

Let me add that:

  • if $2$ is invertible in $R$ then $\langle x+y,x-y\rangle=\langle x,y\rangle$. This follows from $$ x=\frac12((x+y)+(x-y)),\qquad y=\frac12((x+y)-(x-y)). $$
  • if $2$ is not invertible in $R$, but $char(R)\neq2$, then in general the inclusion $\langle x+y,x-y\rangle\subset\langle x,y\rangle$ is not an equality. For instance, in $\Bbb Z$ one has $2\Bbb Z=\langle10,6\rangle\neq\langle16,4\rangle=4\Bbb Z$.
  • if $2=0$ in $R$ then $x+y=x-y$ so that $I+J=I=J$.
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Thank you! .... – Rudy the Reindeer Apr 12 '12 at 12:12

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