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.
 A: 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.
