How do I show that $(\bar x, \bar z)$ is a prime ideal in $Q[x,y,z]/(xy-z^2)$? I proceeded as $\bar a \bar b\in (\bar x, \bar x)\implies \bar a \bar b= r \bar x+ s\bar z\implies ab-rx-sz \in (xy-z^2)$
But I see no way forward from here in showing that one of $\bar a$ or $\bar b$ lies in $(\bar x, \bar z)$
Here ‘bar’ denotes passage from $Q[x,y,z]$ to the quotient $Q[x,y,z]/(xy-z^2)$.
This is a question related to this question here.
Can you tell me if this is right:
I'd like to define a surjective homomorphism from $k[x,y,z]/(xy-z^2)$ onto $k[y]$ with kernel $(\bar{x}, \bar{z})$. To this end I define
$$ f: p(x,y,z) + (xy-z^2) \mapsto p(0,y, 0)$$
Then $f$ is surjective sinc...
Inspired by the idea, I wrote the following solution: $(xy-z^2)$ is an ideal of $(x,z)$. So by the third isomorphism theorem: $Q[x,y,z]/(xy-z^2)/(x,z)/(xy-z^2)\simeq Q[x,y,z]/(x,z)\simeq Q[y]$
Since $(x,z)/(xy-z^2)=(\bar x, \bar z)$, it follows that $(\bar x, \bar z)$ is a prime ideal of Q[x,y,z]/($xy-z^2)$.
The last isomorphism in the second line is achieved by defining a map that is identity of Q[y] and sends x,z to 0.
Suppose that $F$ is a field and let $F[x,y]$ be a polynomial ring. The map $\phi: F[x,y]\to F[y]$, defined by $\phi(f(x,y))=f(y^2,y)$ has kernel $(y^2-x)$. I want to prove that the kernel is as the statement says.
It is clear that $\phi(x)=y^2, \phi(y)=y.$
For any $u(x,y)\in (y^2-x)$, there exist...
huh. well it seems pretty clear that the polynomials a_j don't have to be the zero polynomials. for example (y^2) y^4 + (-y^4) y^2 = 0, but neither y^2 nor -y^4 is the zero polynomial so that just won't work.
i dunno how a geometer would handle this. have you tried dividing your a_n(x) y^n + ... + a_0(x) by y^2 - x? thinking of the division algorithm in K(x)[y]. that would give you a remainder to look at. i dunno.
well, even if F is algebraically closed you wouldn't get what you want from the given argument, because the 'coefficients' you want to be 0 don't come from F, they have ys in them. equating coefficients of all powers of y to 0 gives you just some goofy relations between the coefficients (in F) of the various a_j's.
we went looking for rabbits this afternoon but didn't find any. we sometimes see them in a particular part of our neighborhood, so we went there. only a squirrel.
upvotes and downvotes live in different spaces, one is not the negative of the other.
@Koro even though I know the sound for decades, when I hear foxes when backpacking it always spooks me a bit until I realise that it is a fox and not a human.
there's been discussion of this on meta. partial 'last seen' data is available but it is not very granular. personally i don't think the level of granularity really affects this very much on short time scales.
i probably wouldn't comment on a 5 year old post or answer, but i don't think last seen 'this week' or 'this month' vs. today would affect my behavior, if i bothered to look at that information, which i generally don't.
i'm not active enough to have experienced that but the very occasional unexplained downvote is perplexing. (basically all of my answers are old answers)
i'm reminded of a time SFMOMA had a thing on 'surrealist' painting. a whole floor of the museum was set aside for it. it turns out you only need to see like three or four to get the idea.
the sfmoma cafe (into which i brought food for my kids) had a sign that said "no food is to be consumed inside or outside the cafe". seemed a little threatening from a logical perspective. i have a picture somewhere.
no, it exists. well, i guess it's been described as erotic. i haven't seen it. that isn't my description.
he is partially nude in it. that's all i know. they had a screening or two in downtown berkeley after i left the bay area, and someone i knew emailed me about it.
haha they had some signs like that in rough areas of london the last time i was there. by council estates. roughly: "don't stop here, people will knock you down and take your phone." maybe not that direct, but fairly offensive.
i've never felt too threatened anywhere. i don't really go out at night, though. in my old neighborhood in the east bay, violent things occasionally happened at night.
one of my friends was beat up in a laundromat for his laundry money, which i think meant that people were just looking to beat someone up. that was near my apartment, but it was also in the evening.
the evening is also when people would try to rob the corner store. best to avoid it.
we lived for about a year in the upper floor of a farmhouse outside of ann arbor. that was actually creepier than living in a bad neighborhood in a city. if the downstairs tenant was gone (as he often was, he was there maybe two weekends a month) there was nobody around.
One of my friend in physics department had dropped the (introductory) matrix group course (undergraduate math course). And the reason why he dropped it is because he think that during the course instructor cover Lie algebra and some representation theory but rather instructor focus more on Lie group itself. What's interesting is that my friend didn't even take abstract algebra course (took linear algebra but didn't do very well). I don't know why he's so overestimate himself.
I found that many of physics department friend have some similar manner. I tried to give some advice to him but don't know what to say. I think he doesn't know the importance of base(?)
@MatheusSousa This is very fast: An algorithm for computing logarithms and arctangents, by B. C. Carlson (1972). It's based on a modified arithmetic-geometric mean, so it requires square roots, but all other operations are rational, and it only uses trivial (binary power) divisions. You get better than 5 digits of precision for ln(2) from only 2 rounds.
@PM2Ring Thank you very much :-) So, how should I interprete $f(1)$? As $f(1,...,1)$? Also, what should I think of domain of $f$? I mean $\Bbb R^n$ for which $n$?
@Silent solving the $\nabla f=\frac{\vec r}{r^5}$, I get $f=\frac13-\frac1{3r^3}$, but $f$ is operating on $\mathbb{R^n}$. This has $f\!\left(\vec r\right)=0$ when $r=1$ and $r=\left|\vec r\right|$
One could take $n=1$
I see that PM 2Ring has already clarified this some.
@Silent Well, experience with the gradient of radial functions told me that since the gradient was proportional to $\frac1{r^4}$, the function should be proportional to $\frac1{r^3}$. We need the constant $-\frac13$ to make the result $\frac{\vec r}{r^5}$. The additive constant was to make it $0$ when $r=1$.
$\nabla f(r)=f'(r)\frac{\vec r}{r}$ where $r=\left|\,\vec r\,\right|$
This is the chain rule with $\nabla\left|\,\vec r\,\right|=\frac{\vec r}{\left|\,\vec r\,\right|}$
if the book doesn't offer a definition, it may not be planning on engaging too much with that concept later. 'real' definitions might even be confusing. i don't know.
@leslietownes It does use it later a bit though ... However despite this hiccup with the book I still recommend it :) I feel there have been so many concepts I've been exposed to which are finally making sense
I have stuff on projective geometry in a few sections of the last chapter of my algebra book, but it uses very little algebra — just basic linear algebra.
my daughter's piling a bunch of her clothing on the sofa in my office. running back and forth between my office and her room. she's calling the sofa her 'easter basket.' this is not part of any tradition that i know of.
we regifted her some jelly beans that her mother's aunt gave her the other day. they're in a plastic egg. maybe that's under the pile of clothes. i dunno.
i think you can only ping from chat people who have been in that specific chat. and it may be that you can only ping in comments people who are part of the Q/A chain you are commenting on.
but that's more of an educated guess than knowledge.
@PurpleHaze Yes, you should remove spaces. You can ping anyone who's ever been in that chatroom, but auto-completion only works for people who've visited in the last week or so. Partial names also ping, but be wary of using them because they ping everyone whose name they match.
This is a question related to this question here.
Can you tell me if this is right:
I'd like to define a surjective homomorphism from $k[x,y,z]/(xy-z^2)$ onto $k[y]$ with kernel $(\bar{x}, \bar{z})$. To this end I define
$$ f: p(x,y,z) + (xy-z^2) \mapsto p(0,y, 0)$$
Then $f$ is surjective sinc...
