TITLE: Hilbert function of $k[x_1, x_2, x_3]/ (x_1^2,x_2^2x_3,x_2^3)$
QUESTION [2 upvotes]: If $R = k[x_1, x_2, x_3]$ is a polynomial ring and $I = (x_1^2,x_2^2x_3,x_2^3)$ how do you see that the Hilbert function $H(R/I,i) = 4$ for $i \geq 4$?

So the free resolution of $R/I$ is
$$0 \to R(-6) \to R(-4) \oplus R(-5)^2 \to R(-2) \oplus R(-3)^2 \to R \to R/I \to 0.$$
Now I know that $$H(R(-i),i) = \left\{\begin{matrix}
0 & t < i\\
 \binom{n-1+t-i}{t - i}& t \geq i
\end{matrix}\right.$$
So, for example in $R(-6)$, I should get $\binom{2+t-6}{t-6}$. My problem is how to deal with the direct sums in the other complexes?

REPLY [3 votes]: Why do you need a free resolution in order to find the Hilbert function of this monomial ideal?
If you look at $k[x_1,x_2,x_3]/(x_1^2,x_2^2x_3,x_2^3)$ can immediately notice that the surviving monomials of degree $\ge4$ are: $\{x_3^4,x_2x_3^3, x_1x_3^3,x_1x_2x_3^2\},\{x_3^5,x_2x_3^4,x_1x_3^4,x_1x_2x_3^3\}$, and so on.
However, if you insist to use a resolution for finding $H(R/I,i)$, then notice that $$H(R/I,i)=H(R,i)-H(R,i-2)-2H(R,i-3)+H(R,i-4)+2H(R,i-5)-H(R,i-6)=\binom{i+2}{i}-\binom{i}{i-2}-2\binom{i-1}{i-3}+\binom{i-2}{i-4}+2\binom{i-3}{i-5}-\binom{i-4}{i-6}=4$$ for $i\ge4$.


TITLE: River boat round trip
QUESTION [0 upvotes]: A river boat can travel at $20$ km per hour in still water.  The boat travels $30$ km upstream against the current
          then turns around and travels the same distance back with the current.  If the total trip took $7.5$ hours,
     what is the speed of the current? Solve this question algebraically as well as graphically.

REPLY [0 votes]: $x=(v_{\rm still}-v_{\rm current})t_1$ (when it goes upstream)
$x=(v_{\rm still}+v_{\rm current})t_2$  (when it comes back downstream)
$t_1+t_2=7.5$
and solve for $v_{\rm current}$ , you know the rest of the variables.


TITLE: How are positive intervals calculated?
QUESTION [0 upvotes]: Give these equations
$N=p^4-p^3+161*p$
$N+(n/2)^2=M^2$
$p*(p+n)=161$
How to calculate the intervals in which $n>0$ and $p>0$ and $M>0$ and $N>0$ ?
$n$ and $N$ and $p$ and $M$ not imaginary

REPLY [0 votes]: All you need, in fact, is that $M>\frac{n}{2}$.
First, the $3^{rd}$ line gives us $p^2 + np = 161$. This is always satisfied for $p>0$, $n>0$.
Now, substituting this in the $1^{st}$ line, we get - $$N = p^4 + np^2 = p^2(p^2+n)$$
So, if $n>0$, $N>0$.
Thus, considering the $2^{nd}$ equation, we see that $N>0$ only if $M>\frac{n}{2}$.