Proving R[x_1...x_n]/(x_1-r_1...x_n-r_n) isomorphic to R

[frosty mural]

It's tempting to find $\ker f : R[x_1,...,x_n] \rightarrow R$ but I don't see a way to prove that the kernel is $(x_1-r_1,...,x_n-r_n)$ from previous work

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[scenic notch]

muffin

[frosty mural]

Could I do a change of variables from $R[x]/(x-r) \cong R$, letting $r = r_1 + ... + r_n$? I feel as though that is not enough

[scenic notch]

muffin

[pulsar seal]

What does 5.41 refer to?

But you can just note that the evaluation homeomorphism should have kernel that ideal

If it's in the ideal, then clearly it evaluates to 0 at (r1,...,rn)

[pulsar seal]

Right

so yeah, just define the evaluation homomorphism as $f(x_1,...,x_n)\mapsto f(r_1,...,r_n)$

[scenic notch]

Omegabet_

[pulsar seal]

Then $\ker(\text{ev})=(x_i-r_i\mid 1\leq i\leq n)$

[scenic notch]

Omegabet_

[pulsar seal]

$\supset$ is clear, for $\subset$, apply the multivariate polynomial division

[scenic notch]

Omegabet_

[pulsar seal]

you get some $f(x_1,...,x_n)=\sum_{i=1}^n (x_1-r_1)q(x_1,...,x_n)+r$ where $r\in R$

[scenic notch]

Omegabet_

[pulsar seal]

since $f$ evaluates to $0$ at $(r_1,...,r_n)$, you get $r=0$. Alternatively, you can probably do some inductive argument

[scenic notch]

Omegabet_

[frosty mural]

This makes a lot of sense. Thanks!

+close