# How to show the ideal $\mathfrak a$ is not finitely generated?

Let $k$ be a field and let $A=k\{x,y\}$ be the free algebra over $k$ on two generators. Let $z_i=yx^i$. Consider the left ideal $\mathfrak a=\sum_{i\geqslant 0} Az_i$. How can one show $\mathfrak a$ is not finitely generated?

I encountered this while trying to exhibit a ring $A$ and a short exact sequence of left $A$-modules whose last two terms where finitely generated, but whose first term wasn’t.

#### Solutions Collecting From Web of "How to show the ideal $\mathfrak a$ is not finitely generated?"

One way to do this is to proceed as follows:

• first, your algebra is graded, with $x$ and $y$ in degree $1$, and your ideal is a graded ideal, that is, it is generated by homogeneous elements.
• If $M$ is a finitely generated graded module over a non-negatively graded algebra algebra $A$ with $A_0$ a field, and we write $A_+$ the ideal generated by positively graded elements, then the vector space $M/A_+M$ has dimension equal to the cardinal of a minimal set of generators.
• In your case, $M/A_+M$ is infinite-dimensional.

Note that any element of the ideal is sum of elements ending in $yx^i$ for various $i$s. Now suppose the ideal was finitely generated. This means there is a largest $i$ occurring in these sums: since the algebra is free, any two different words are linearly independent and hence we cannot obtain higher occurrences of powers of $x$ to the right. But in our ideal we have arbitrarily large powers of $x$ to the right, so it must be the case the ideal not finitely generated.