@Alizter A ring $R$ is called Noetherian if every ascending chain of ideals eventually stabilize, i.e., if $I_n$ are ideals of $R$ and there is no infinitely long ascending chain $I_0 \subset I_1 \subset I_2 \subset I_3 \subset I4 \subset \, \cdots$. Prove that if $R[X]$ is Noetherian so is $R$.
Is there a simpler way of finding an expression for this? $\displaystyle \bigcup_{x \in [0,1]} [x,1] \times [0,x^2] = \{ (x,y) : x \in [f,1], y \in [0,f^2] \text{ for } f \in [0,1] \}$
@r9m 1. $\angle A=80^{0}, AI+IB=BC$, $I$ incenter, find $\angle B$. 2. $al_a=bl_b=cl_c$ such that $l_a$, $l_b$, $l_c$ are angle bisectors of $\angle A$, $\angle B$... Prove its an equilateral $\triangle$.
Is there a simpler way of finding an expression for this? $\displaystyle \bigcup_{x \in [0,1]} [x,1] \times [0,x^2] = \{ (x,y) : x \in [f,1], y \in [0,f^2] \text{ for } f \in [0,1] \}$
I know that it looks like $\displaystyle \int_{0}^{1} x^2 \text{ d}x$, but I can't find a nice way of expressing the union in a simpler way, @BalarkaSen.
@IceBoy A much more geometrically sound construction is to draw parallel lines on a sheet of page (that is your reals) and identify the four corners by a bit of glue.
@BalarkaSen Clearly $R\subset R[X]$ therefore if $R[X]$ has a maximal ideal $K$ then the maximal ideal $I$ of $R$ is $I\subset K$ therefore $R$ has a maximal ideal and is noetherian.
If $\displaystyle \bigcup_{\alpha \in I} A_{\alpha} = \bigcap_{\alpha \in I} A_{\alpha}$ for some index set $I$, is it sensible to conclude that all the $A_{\alpha}$ are equal?
@BalarkaSen Clearly $R\subset R[X]$ therefore if $R[X]$ has a maximal ideal $K$ then the maximal ideal $I$ of $R$ is $I\subset K$ therefore $R$ has a maximal ideal and is noetherian.
@Alizter You need ascending chain conditions instead of tweaking and twiddling with maximal elements. Hint : Prove that if $R$ is Noether, so is $R/I$ for some ideal $I$.
@Alizter Wait, I don't get it. How do you even know that $I$ exists?
