Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $I$ is an ideal in a polynomial ring $R=k[x,y]$. Let $\overline{k}$ be the algebraic closure of $k$ and let $S=\overline{k} [x,y]$. Then is $IS\cap R=I$?

share|improve this question
    
Isn't $R\subset S$ faithfully flat? –  user26857 Mar 2 at 19:43

2 Answers 2

All rings will be commutative.

Lemma 1 Let $K$ be a field. Let $L$ be a non-zero $K$-algebra. Let $N$ a $K$-module. Then the canonical homomorphism $N \rightarrow N\otimes_K L$ sending $x$ to $x\otimes 1$ is injective.

Proof: Suppose $x \neq 0$. There exists a basis of $N$ over $K$ containing $x$. Since $1 \neq 0$ in $L$, there exists a basis of $L$ over $K$ containing $1$. Hence there exists a basis of $N\otimes_K L$ over $K$ containing $x\otimes 1$. Hence $x\otimes 1 \neq 0$ QED

Lemma 2 Let $K$ be a field. Let $A$ a $K$-algebra. Let $L$ be a non-zero $K$-algebra. Let $B = A\otimes_K L$. Then the canonical homomorphism $M \rightarrow M\otimes_A B$ is injective for any $A$-module $M$.

Proof: $M\otimes_A B = M\otimes_K L$. Hence the assertion follows from Lemma 1. QED

Proposition Let $K$ be a field. Let $A$ be a $K$-algebra. Let $L$ be a non-zero $K$-algebra. Let $B = A\otimes_K L$. By Lemma 1, we can identify $A$ as a subring of $B$ by the canonical homomorphism $A \rightarrow A\otimes_K L$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: By Lemma 2, the canonical homomophism $A/I \rightarrow A/I\otimes_A B$ is injective. Since $A/I\otimes_A B = B/IB$, we are done. QED

Corollary Let $K$ be a field. Let $A = K[X_1,\dots, X_n]$ be a polynomial ring. Let $L$ be an extension field of $K$. Let $B = L[X_1,\dots, X_n]$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: Since $B = A\otimes_K L$, the assertion follows immediately from the proposition. QED

share|improve this answer

Lemma Let $K$ be a field. Let $L/K$ be an extension field. Let $K[X_1,\dots, X_n]$ and $L[X_1,\dots, X_n]$ be polynomial rings. Let ($\omega_i$) be a linear basis of $L$ over $K$. Then every element $f \in L[X_1,\dots, X_n]$ can be uniquely written as $f = \Sigma_i \omega_i f_i(X)$, where $f_i(X) \in K[X_1,\dots, X_n]$.

Proof: Let ($M_{\alpha}$) be the family of all the monomials of $K[X_1,\dots, X_n]$. Let $f = \Sigma_{\alpha} c_{\alpha}M_{\alpha}$, where $c_{\alpha} \in L$. Let $c_{\alpha} = \Sigma_i a_{\alpha i} \omega_i$, where $a_{\alpha i} \in K$. Then $f = \Sigma_{\alpha} \Sigma_i a_{\alpha i} \omega_i M_{\alpha} = \Sigma_i \Sigma_{\alpha} a_{\alpha i} M_{\alpha} \omega_i = \Sigma_i \omega_i f_i(X)$, where $f_i(X) = \Sigma_{\alpha} a_{\alpha i} M_{\alpha}$.

Next we prove the uniqueness. Suppose $\Sigma_i \omega_i f_i(X) = 0$, where $f_i(X) \in K[X_1,\dots, X_n]$. Suppose $f_i(X) = \Sigma_{\alpha} a_{\alpha i} M_{\alpha}$, where $a_{\alpha i} \in K$. Then $\Sigma_i \omega_i f_i(X) = \Sigma_i \Sigma_{\alpha} \omega_i a_{\alpha i} M_{\alpha} = \Sigma_{\alpha} \Sigma_i \omega_i a_{\alpha i} M_{\alpha} = 0$. Hence $\Sigma_i \omega_i a_{\alpha i} = 0$ for each $\alpha$. Hence $a_{\alpha i} = 0$. Hence $f_i(X) = 0$. QED

Proposition Let $K$ be a field. Let $A = K[X_1,\dots, X_n]$ be a polynomial ring. Let $L/K$ be an extension field. Let $B = L[X_1,\dots, X_n]$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: Let $f \in IB$. Let ($\omega_i$) be a linear basis of $L$ over $K$. We can assume that one of $\omega_i$, say $\omega_{i_0}$ is 1. Let $f_1,\dots,f_n$ be generators of $I$. We can write $f = \Sigma_k g_k f_k$, where $g_k \in B$. Let ($M_{\alpha}$) be the family of all the monomials of $K[X_1,\dots, X_n]$. Suppose $g_k = \Sigma_{\alpha} c_{\alpha k} M_{\alpha}$, where $c_{\alpha k} \in L$. Then $f = \Sigma_k g_k f_k = \Sigma_k \Sigma_{\alpha} c_{\alpha k} M_{\alpha} f_k$. Since $M_{\alpha} f_k \in I$, we can write $f = \Sigma_i \omega_i h_i$, where $h_i \in I$. Suppose $f \in IB \cap A$. By Lemma, $f = h_{i_0}$. Hence $f \in I$. QED

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.