# Isn't a subalgebra of a finitely generated k-algebra always finitely generated? [duplicate]

Let $K[x_1, x_2,\dots, x_n]$ be a polynomial ring. If it is a graded ring, then under certain conditions, its subalgebras may be finitely generated.

Isn't a subalgebra of a finitely generated k-algebra always finitely generated?

## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 16 '16 at 17:21

• No. Consider the free $k$-algebra $k\langle x,y\rangle$, say. – Pedro Tamaroff Jan 16 '16 at 16:14
This is false even in the commutative case. For example, $k[x, y]$ is finitely generated, but it has a subalgebra $k[x, xy, xy^2, xy^3, \dots ]$ which is not (this is a nice exercise).
• I'm sorry I'm having trouble understanding this. Isn't $k [x, xy, xy^2, xy^3, \dots]=k [x]$? – fierydemon Jan 16 '16 at 16:56
• @Ayush: no. $k[x]$ doesn't contain $xy$. This is a subalgebra, not an ideal. – Qiaochu Yuan Jan 16 '16 at 16:58