Some basic question about a quotient ring

1. May 10, 2012

naturemath

It's been awhile since I studied ring theory but here's a question about it:

Let R = C[x1, x2, x3, x4, x5, x6, x7, x8] be a complex polynomial ring in 8 variables.

Let

f1 = x1 x3 +x5 x7 and
f2 = x2 x4 +x6 x8.

How do $\bar{f}$1, $\bar{f}$2 in (f1,f2)R/(x1,x2)R look like?

Is it
$\bar{f}$1 = x5 x7 + (x1,x2)

$\bar{f}$2 = x6 x8 + (x1,x2)

or is it

$\bar{f}$1 = x1 x3 + x5 x7 +(x1,x2)

$\bar{f}$2 = x2 x4 + x6 x8 + (x1,x2)?

Last edited: May 10, 2012
2. May 10, 2012

naturemath

Since

x1*x3 $\in$ (x1, x2),

x1 x3+ x5 x7 ~ x1 x3 mod (x1, x2).

So the coset x1 x3+ x5 x7 + (x1, x2) = x5 x7 + (x1, x2)?

3. May 10, 2012

DonAntonio

What is that?? What do you mean by $\,\,(f_1,f_2)R/(x_1,x_2)R\,\,$? The denominator clearly is

an ideal, but what is the numerator?

DonAntonio

4. May 10, 2012

naturemath

> (f1,f2)R/(x1,x2)R?

The numerator (f1, f2) is an ideal in R generated by f1 and f2 while the denominator (x1, x2) is an ideal in R generated by x1 and x2.

5. May 10, 2012

naturemath

So what I'm interested in is: how do the images of f1 and f2 look like in (f1,f2)R/(x1,x2)R?

6. May 10, 2012

naturemath

It seems to me that in (f1,f2)R/(x1,x2)R, we are killing off all polynomial combination of x1 and x2 (all polys of the form ax1 + bx2 where a and b are in R), or are we killing off something smaller than that?

Thus, do

$\bar{f}$1 = x5 x7

$\bar{f}$2 = x6 x8

in (f1,f2)R/(x1,x2)R ?

7. May 10, 2012

mathwonk

yes. that quotient ring just cancels all terms containing x1 or x2.

8. May 12, 2012

naturemath

So (f1,f2)R/(x1,x2)R is generated by x5 x7 and x6 x8, right?

I.e., (f1,f2)R/(x1,x2)R = (x5 x7, x6 x8)