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06:38
Hello! I want to ask: what should I know so that I can answer this: Is the ideal I=(X+Y,X−Y) in the polynomial ring C[X,Y] a prime ideal?
To answer this, I have the following two ideas, are these enough to answer the above question?:
1) One way to solve this of course would be to solve it by definition i.e., taking f(x,y) and g(x,y) from C[x,y] and then noting that there exist p(x,y) and q(x,y) in C[x,y] such that f(x,y)g(x,y)=p(x,y)(x+y)+q(x,y)(x-y) and then concluding from here that either p(x,y) or q(x,y) is in (x+y,x-y). But this seems very lengthy too me.
07:17
So that if somebody answers here, they might check also what was posted there. (To avoid duplication.)
5
Q: Is the ideal $I=(X+Y,X-Y)$ in the polynomial ring $\mathbb{C}[X,Y]$ a prime ideal?

Subhasis BiswasIs the ideal $I=(X+Y,X-Y)$ in the polynomial ring $\mathbb{C}[X,Y]$ a prime ideal? Justify your answer. I am a bit hesitant about asking this here. The question is not "How to Solve This Problem". The question is "What do I need to learn first in order to solve this type of problems". I have ...

1
Q: Is the ideal $R= (x+y, x−y)$ in the polynomial ring $\mathbb{C}[x, y]$ a prime ideal?

Avijit DikeyI wanted to check if my approach is correct: Consider : $$\phi:\mathbb{C}[x, y] \rightarrow \mathbb{C}[x]$$ $$ y \mapsto -x$$ $$x \mapsto x$$ is a surjective homomorphism, with $\ker(\phi)=(x+y)$. Hence, $\mathbb{C}[x, y]/(x+y) \cong \mathbb{C}[x]$. Now, I use the third isomorphism theorem to as...

07:29
@Koro Wouldn't the suggestion in these comments lead to a relatively straightforward solution?
$\Bbb C [x]/(2x)$ is isomorphic to $\Bbb C$, which is a field. Anyway, you can easily see that $(x+y, x-y) = (x,y)$ which is obviously a maximal ideal. — Crostul Aug 27, 2020 at 7:36
@Crostul yes of course, silly me. $2$ is invertible over $\mathbb{C}$. What if the question were modified to $\mathbb{Z}[x]$? Here, $2$ is not invertible and hence $(2x) \neq (x)$. — Avijit Dikey Aug 27, 2020 at 9:20
You are right: over $\Bbb Z$ it wouldn't be a prime ideal. — Crostul Aug 27, 2020 at 10:06
A similar comment is also on another post:
I agree with Feng Shao that you should know a little bit of machinery, and in particular your lacking of understanding for $\mathbb{C}[X,Y]$ is an indicator of the act that you should improve your knowledge! $$\;$$ However, my approach would be: prove that $(X+Y,X-Y) =(X,Y)$ and then use Nullstellensatz, and you are done, as it will be even maximal. However, by filling in the blanks here, you will need to understand and learn/practice algebra — Enkidu May 20, 2019 at 13:01
@MartinSleziak thanks for the response! I'm actually new to fields and I would also like to mention I don't yet know Nullstellensatz theorem.
I'll go through the comments you shared. Thanks!
I believe that since $2x=(x+y)+(x-y)$, it follows that $2x \in (x+y,x-y)$ and similarly $2y\in (x+y,x-y)$ and since $2$ is a unit in C, it follows that $x,y\in (x+y,x-y)$. So $(x,y)\subset (x+y,x-y)$. And it is clear that: $x+y,x-y\in (x,y)$ so $(x+y,x-y)\subset (x,y)$. It follows that $(x+y,x-y)=(x,y)$.
Now, the problem is to analyze $C[x,y]/(x,y)$
I do not think that you need Nullstellensatz. (I guess that might be helpful if you wanted to find all maximal ideals.)
Well, one possibility would be to show that $\mathbb C[x,y]/(x,y)\cong\mathbb C$.
But I'd guess that you could show that $(x,y)$ is a maximal ideal directly from the definition, too.
I'd like to work on the former: showing that C[x,y]/(x,y)$\simeq \mathbb C$. I'll try to find a ring homomorphism from C[x,y] to C with kernel (x,y).
2
Q: This ideal is prime. But is it also maximal?

user16008Could someone please give me a hint how to prove (preferably directly, without finding clever homomorphisms) that $(X,Y)$ is a maximal ideal in $\mathbb{C}[X,Y]$ ? This ideal is prime, since it contains all polynomials of all degree that don't have a constant term, so my guess was, it's maybe al...

I'll have to leave - I have rather busy day today.
07:45
@MartinSleziak thank you so much! I think I got my answer now. That was a very 'tricky' isomorphism-mapping polynomials to their constant terms.
@MartinSleziak Bye and have a good day ahead. Thanks again for the help!
 
3 hours later…
11:03
I want to ask one question. I'm trying to show that group of order 1320 is simple. To do this suppose $G$ is simple group of order 1320. Then by Sylow theorem, the number of sylow $11$-subgroup is $12$. Hence there is an induced monomorphism $G\to S_{12}$. I want to show this actually factor through $A_{11}$.

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