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Let $X$ be an affine algebraic set and $f\in K[X]$ where $K[X]$ is the coordinate ring of $X$. Suppose $I(X)=\langle G_1,\ldots,G_r\rangle$ and $W=Z(G_1,\ldots,G_r,FT_{n+1}-1)$, where $G_1,\ldots,G_r\in k[T_1,\ldots,T_n]$.

The image of $F\in k[T_1,\ldots,T_n]$ in the quotient $K[X]$ is $f$, i.e, $f=F+I(X)$.

I'm trying to prove these isomorphisms:

$$k[W]\cong \frac{k[T_1,\ldots,T_n,T_{n+1}]}{\langle G_1,\ldots,G_r,FT_{n+1}-1\rangle}\cong \frac{k[X][T_{n+1}]}{\langle fT_{n+1}-1\rangle}\cong k[X][1/f].$$

I couldn't prove it (I tried a lot), if someone could help me with some of these isomorphisms it would be very helpful to me and I would be very grateful.

I really need help.

Thanks a lot

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  • $\begingroup$ Where do you still have problem? $\endgroup$ – Ehsan M. Kermani Dec 6 '13 at 7:14
  • $\begingroup$ Dear @EhsanM.Kermani I still don't completely understand why $(k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)) [T_{n+1}] \cong k[T_1,\cdots,T_n][T_{n+1}]/(G_1,\cdots,G_r)$ and why the isomorphism is preserved if you take the quotient from both sides by the ideal $(fT_{n+1}-1)$. Thank you. $\endgroup$ – user75086 Dec 6 '13 at 9:29
  • $\begingroup$ Dear @user75086, I added the detailed explanation. Usually, you should be able to fill all the gaps when you see the guidelines. If you see you can't, this indicates you've missed some preliminary materials. As an advice, going through them helps a lot. Cheers! $\endgroup$ – Ehsan M. Kermani Dec 6 '13 at 22:28
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Well, break it down into some small pieces as follows: $k[X]=k[T_1,\cdots,T_n]/(G_1,\cdots,G_r),$ so

$$k[X][T_{n+1}]=(k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)) [T_{n+1}] \cong k[T_1,\cdots,T_n][T_{n+1}]/(G_1,\cdots,G_r)$$

which is in fact

$$k[X][T_{n+1}]\cong k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r).$$

Then show that the isomorphism is preserved if you take the quotient from both sides by the ideal $(fT_{n+1}-1).$

For the final isomorphism, let $R=k[X],$ define the ring morphism $R[T_{n+1}] \to R[1/f]$ by sending $T_{n+1} \mapsto 1/f,$ then show that it is surjective and its kernel is the ideal $(fT_{n+1}-1).$

Added:

  • Building the first isomorphism.

To prove $R[X]/I[X] \cong (R/I)[X]$ for $R=k[T_1.\cdots,T_n]$ and $I=(G_1,\cdots,G_r)$ define $g: R[X] \to (R/I)[X]$ by sending a polynomial with coefficient in $R$ to the polynomial with coefficient in $R/I.$ That is, first consider the projection morphism $\pi: R \to R/I$ sending $r \mapsto r+I$ and then

$$g(r_nX^n+\cdots+r_1X+r_0)=(r_n+I)X^n\cdots+(r_1+I)X+(r_0+I)$$

It is easy to show that $g$ is a surjective ring homomorphism. Basically $g$ is constructed from $\pi$ which is a surjective ring homomorphism. Now, in order to use the first isomorphism theorem for rings we should show that the kernel of $g$ is $I[X],$ which is clear since the kernel of $\pi$ is $I$ and $g$ was constructed from $\pi$ by adding a variable $X,$ so the kernel of $g$ is $I[X],$ or if $g(r_nX^n+\cdots+r_1X+r_0)=0$ means that $(r_n+I)X^n\cdots+(r_1+I)X+(r_0+I)=0$ and $0$ in the ring $(R/I)[X]$ is $I[X]$ because a polynomial is equivalent to zero (always zero) if and only if all of its coefficients are zero, that is, $r_i+I$ is zero in $R/I$ for $1 \leq i \leq n,$ and the zero element in $R/I$ is just $I,$ so $r_i+I=I$ implying $r_i \in I$ for all $i.$

Remark: As I explained in my last comment, since $I=(G,\cdots,G_r)$ is an ideal of $k[T_1,\cdots,T_n]$ then $I[T_{n+1}]$ will be an ideal in $k[T_1,\cdots,T_n,T_{n+1}],$ but we still can write $(G,\cdots,G_r)$ instead of $(G,\cdots,G_r)[T_{n+1}]$ when we say consider $(G,\cdots,G_r)$ now as an ideal in $k[T_1,\cdots,T_n,T_{n+1}]$ with no harm.

  • Why taking quotient of $(fT_{n+1}-1)$ is preserving the isomorphism?

Back to our notation, we showed that

$$k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \cong (k[T_1,\cdots,T_n]/(G_1,\cdots,G_r))[T_{n+1}]$$

and since $k[X]=k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)$ we then have

$$\varphi: k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \cong k[X][T_{n+1}]$$

say the isomorphism map is called $\varphi.$

To prove

$$k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r,FT_{n+1}-1) \cong k[X][T_{n+1}]/(fT_{n+1}-1)$$

again define a map

$$p: k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \to k[X][T_{n+1}]/(fT_{n+1}-1)$$

by sending an element in $s \in k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r)$ to $\varphi (s) \in k[X][T_{n+1}]$ and the sending it to $\pi (\varphi (s)) \in k[X][T_{n+1}]/(fT_{n+1}-1)$ where

$$\pi: k[X][T_{n+1}] \to k[X][T_{n+1}]/(fT_{n+1}-1)$$

is the natural projection.

Since $p=\pi \circ \varphi$ then is a ring homomorphism and since $\pi$ is surjective and $\varphi$ is an isomorphism then their composition $\pi \circ \varphi$ is surjective, so $p$ is a surjective ring homomorphism. Again to use the first isomorphism theorem for rings, we should show that the kernel is $(FT_{n+1}-1)$ which is clear since $p(s)=0$ for some $s \in k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r)$ is meaning that $\pi (\varphi(s))=0$ which means $\varphi(s) \in \ker(\pi)=(fT_{n+1}-1).$ So $\varphi(s)=(fT_{n+1}-1)t$ for some $t \in k[X][T_{n+1}]$ by definition of an ideal. Since $\varphi$ is an isomorphism so its inverse is also an isomorphism then $s=\varphi^{-1}((fT_{n+1}-1)t)=\varphi^{-1}(fT_{n+1}-1) \varphi^{-1}(t).$ The way $\varphi$ was constructed shows that $\varphi^{-1}((fT_{n+1}-1))=FT_{n+1}-1$ because $F \mapsto f$ via the projection $k[T_1,\cdots,T_n] \to k[X].$ Therefore, $s \in (FT_{n+1}-1)$ and $\ker p \subset (FT_{n+1}-1).$ Obviously $(FT_{n+1}-1) \subset \ker p,$ hence the equality, so we're done.

For the last isomorphism in the question follow the same approach and the last paragraph before "Added."

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  • $\begingroup$ Dear Ehsan, thank you very much for your answer. I couldn't prove that $(k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)) [T_{n+1}] \cong k[T_1,\cdots,T_n][T_{n+1}]/(G_1,\cdots,G_r)$, can you give me some hints how to proceed? $\endgroup$ – user75086 Dec 4 '13 at 8:31
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    $\begingroup$ @user75086, Basically, you want to prove that $(R/I)[x] \cong R[x]/I.$ Note that you can write any polynomial with coefficient in $R/I,$ as follow: $(a_n+I)x^n+\cdots+(a_1+I)x+(a_0+I)=(a_nx^n+\cdots+a_1x+a_0)+I.$ $\endgroup$ – Ehsan M. Kermani Dec 4 '13 at 19:14
  • $\begingroup$ Then they are equal, not only isomorphic? $\endgroup$ – user75086 Dec 5 '13 at 2:06
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    $\begingroup$ @user75086, No, I wouldn't say that. They're basically two different sets. Also note that, when we write $R/I$ it means $I$ is an ideal of $R$ and so $R[x]/J$ means $J$ is an ideal of $R[x],$ so if you don't want to use abuse of notation, since you might get confused, it's better to write the above isomorphism as $(R/I)[x] \cong R[x]/I[x]$ which would be the same as $R[x]/I$ but now this $I$ is different from the previous $I$ in $R/I.$ $\endgroup$ – Ehsan M. Kermani Dec 6 '13 at 2:33
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    $\begingroup$ Yes, I'm glad you understood it :) Btw the second approach (my last comment) is somewhat easier but less illuminating. $\endgroup$ – Ehsan M. Kermani Dec 10 '13 at 23:24

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