MHB Is $C[x,y,z]/<y-x^2,z-x^3>$ an Integral Domain?

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Domain Integral
Click For Summary
The discussion centers on whether the quotient ring $C[x,y,z]/<y-x^2,z-x^3>$ is an integral domain. It is argued that the kernel of the homomorphism is generated by the elements $y-x^2$ and $z-x^3$. By the isomorphism theorem, the quotient ring is isomorphic to the image of this homomorphism, which is a subring of $\mathbb{C}[x]$. Since $\mathbb{C}[x]$ is an integral domain, the image must also be an integral domain. However, a participant notes that a morphism must be defined to substantiate this claim.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hi! (Smile)

Can I justify like that, that $C[x,y,z]/<y-x^2,z-x^3>$ is an integral domain?

We show that $ker(\phi)=<y-x^2,z-x^3>$.

From the theorem of isomorphism, we have that $C[x,y,z]/<y-x^2,z-x^3> \cong im(\phi)$

$im(\phi)$ is a subring of $\mathbb{C}[x]$

$\mathbb{C}[x]$ is an integral domain, since $\mathbb{C}$ is a field.

As $im(\phi)$ is a subring of $\mathbb{C}[x]$, $im(\phi)$ is also an integral domain.

(Thinking)
 
Physics news on Phys.org
Hi evinda,

You need to define some morphism to claim such things.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 26 ·
Replies
26
Views
821
  • · Replies 5 ·
Replies
5
Views
915
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
6
Views
2K