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I am reading Gortz and Wedhorn's Algebraic Geometry. In their section on the symmetric algebra they explain the adjoint situation $$ \mathrm{Sym}_A \dashv i_A\colon \mathrm{Alg}(A)\to \mathrm{Mod}(A)$$ where $\mathrm{Sym}_A\colon M \mapsto \bigoplus_{i=0}^\infty T^n(M)/{\sim}$, with $T^n(M)=M\otimes_A\cdots \otimes_A M$ and $\sim$ is the equivalence relation generated by $x\otimes y-y\otimes x$ and $i_A$ is just the inclusion.

In passing they say given any ring morphism $f\colon A\to B$ this adjunction gives us the isomorphism $$ \mathrm{Sym}_A(M)\otimes_A B \cong \mathrm{Sym}_B(M\otimes_A B). $$ In other words, the formation of symmetric algebras commutes with extension of scalars.

The proof I have uses the adjointness of the extension and restriction of scalars to arrive at $$ \mathrm{Alg}(B)(\mathrm{Sym}_B(M\otimes_A B),R)\cong \mathrm{Alg}(B)(\mathrm{Sym}_A(M)\otimes_AB, R) $$ for any test object $R$. Hence by Yoneda, we have our iso.

My question: Do we need to use the restriction and extension of scalars adjoint properties to arrive at the solution? It seems the authors suggested we could verify it simply by the $\mathrm{Sym}\dashv i$ adjoint.

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    $\begingroup$ That's how I would do it. Sometimes people take some facts as obvious. $\endgroup$ – Zhen Lin Aug 28 '14 at 15:31
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    $\begingroup$ The same question has been asked before (currently I cannot find it), but my answer also "only" contained the Yoneda proof. Of course, this is just playing around with functors and their adjoints. $\endgroup$ – Martin Brandenburg Sep 3 '14 at 12:43

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