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1d
accepted The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
2d
answered The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
2d
comment Are intermediate rings of finitely generated ring extensions also finitely generated?
For future reference, the proof of the fact that the subring $R\subseteq k[x, y]$ is not finitely-generated over $k$ can be found in this thread.
2d
comment The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
Thanks, this is excellent!
2d
comment The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
@N.H. Not at all! Thanks for the heads-up, because the post was unclear :)
2d
revised The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
clarified the question
2d
comment The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
@N.H. Thanks. I clarified the post. I meant finitely-generated as a $k$-algebra. In other words, the question is, why can't we find finitely many $f_1, f_2, …, f_n\in R$ such that $R = k[f_1, f_2, …, f_n]$? And $R$ is not generated by $x$, because the $k$-algebra generated by $x$ is simply $k[x]$, not $R$.
2d
asked The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$
Jun
26
comment Zero divisor in $R[x]$
Bill, what an elegant proof! Thanks for sharing this lovely argument.
Jun
25
accepted Borel measurability is a local property
Jun
25
answered Borel measurability is a local property
Jun
24
comment Is the ring of Laurent polynomials in $n$ noncommuting variables Noetherian?
Is it true that the polynomial ring $S=R\langle x_1, x_2, …, x_{n} \rangle$ in $n$ non-commuting variables Noetherian ring?
Jun
23
comment Showing a polynomial is irreducible over $\mathbb{C}[x,y]$
@user26857: You are absolutely right: $\mathbb{C}[y]/(y-1)\cong \mathbb{C}$ is a domain, so $y-1$ is prime.
Jun
23
answered Showing a polynomial is irreducible over $\mathbb{C}[x,y]$
Jun
22
comment Show that any quadric in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$
@Leafar: Thanks a lot! Much appreciated.
Jun
22
accepted The natural map $R\mapsto R[t]/(ft-1)$ is injective
Jun
22
comment example of a open function such that the restriction is not open
@Ryoma: If you feel that this answer has satisfied your question, it is encouraged that you accept the answer. Since you haven't accepted any answers before (as of now), I thought I'd point out this link for your information.
Jun
22
comment Show that any quadric in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$
Dear Leafer, could you let me know the source of this problem (textbook or online notes)? I am asking because I am interested in doing problems like this to improve my knowledge about projective geometry. Thanks!
Jun
21
revised The natural map $R\mapsto R[t]/(ft-1)$ is injective
corrected the problem statement
Jun
21
comment Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
Seems to be special case of this MO question, but perhaps there is simpler argument.