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I assume it's an ideal that generates the whole Ring.

Also is $(x+y)^2$ a non cyclic ideal of $R=\mathbb{C}[x,y]/(x^2,xy,y^2)$.

The notes I'm using don't give a definition of a cyclic ideal, assume it's the same as definition of cyclic group but with group replaced with ring.

Found the definition, now. An ideal M is cyclic if $M=\{rm: r \in R \}$ for $m \in M$.

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1 Answer 1

This link suggests it means principal ideal: http://books.google.de/books?id=RjkWZs-6zg8C&pg=PA19&lpg=PA19&dq=%22cyclic+ideal%22&source=bl&ots=U1eT61QUAk&sig=INCuWZWxulLSZEyxTpmzVcnm7CI&hl=de&ei=83m7TuCRI4Ol8QOz8uDZBw&sa=X&oi=book_result&ct=result&resnum=6&ved=0CFYQ6AEwBQ#v=onepage&q=%22cyclic%20ideal%22&f=false I.e. it is an ideal generated by one element.

You usually call a module generates by one element cyclic. Since ideals are also submodules of a ring, the same terminology may be applied to them.

Edit: your ideal, however, seems to be principal, since it is the square of a principal ideal. In fact, if I am not mistaken, $(x+y)^2$ is the zero ideal in $\mathbb{C}[x,y]/(x^2,xy,y^2)$...

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Is (x+y) an principal ideal? –  simplicity Nov 10 '11 at 7:41
    
Well, if $r$ is a ring element (in a ring $R$) then $(r)$ means the ideal generated by $r$, i.e. $\{a\cdot r ~|~ a\in R\}$. In this case we have $(x+y)$, which is the ideal generated by the one element $x+y$. –  Oliver Braun Nov 10 '11 at 7:48

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