# Embedding of free $R$-algebras

Let $R$ be any nontrivial commutative unital ring and $I$ and $J$ any sets with $|I|>|J|$. Does there exist an embedding of $R$-algebras $R[x_i; i\in I]\longrightarrow R[y_j;j\in J]$?

When $R$ is a domain, the question has been answered negatively here, via transcendence degrees.

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No!

Lemma 1 Let $A\to B$ be injective ring map and $\mathfrak{p}$ a minimal prime ideal of $A$. Then there exists a prime ideal $\mathfrak{q}$ of $B$ such that $\mathfrak{q}\cap A=\mathfrak{p}$.

Proof. $A_{\mathfrak{p}}\to B_{\mathfrak{p}}$ is injective. Pick any prime ideal $\mathfrak{q}B_{\mathfrak{p}}$ of $B_{\mathfrak{p}}$. Then $\mathfrak{q}\cap A=\mathfrak{p}$.

Lemma 2 The minimal prime ideals of $A[X]$ are of the form $\mathfrak{p}[X]$, where $\mathfrak{p}$ is a minimal prime of $A$. Here $X$ may stand for infinitely many indeterminates.

Now suppose there exists an injective $R$-algebra map $R[X_i]\to R[Y_j]$. Let $\mathfrak{p}[X_i]$ be a minimal prime of $R[X_i]$, then by Lemma 1, there is a prime $\mathfrak{q}$ of $R[Y_j]$ which contracts to $\mathfrak{p}[X_i]$. So $\mathfrak{q}\supseteq\mathfrak{p}[Y_j]$. We may let $\mathfrak{q}=\mathfrak{p}[Y_j]$. OK, hence we have an injective map $R/\mathfrak{p}[X_i]\to R/\mathfrak{p} [Y_j]$. By comparing the transcendence degrees we know that this is impossible.

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Can you explain the step "We may let $\mathfrak{q}=\mathfrak{p}[Y_j]$"? –  Martin Brandenburg Mar 16 '13 at 13:52
Since $\mathfrak{p}[X_i]$ is a minimal prime, and $\mathfrak{p}[Y_j]\cap R[X_i]\subset \mathfrak{p}[X_i]$, so we can assume $\mathfrak{q}=\mathfrak{p}[Y_j]$. –  wxu Mar 16 '13 at 15:40
@YACP, if not injectve, $B_{\mathfrak{p}}$ maybe is a zero ring! –  wxu Mar 16 '13 at 16:15
I don't see why $p[Y_j] \cap R[X_i] \subseteq p[X_i]$ should hold. –  Martin Brandenburg Mar 16 '13 at 16:43
@MartinBrandenburg $\mathfrak{p}[Y_j]\cap R[X_i]\subseteq \mathfrak{q}\cap R[X_i]=\mathfrak{p}[X_i]$... .. :) –  wxu Mar 16 '13 at 16:46