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For n1n\ge 1, determinant is a natural transformation det:GLn()()×\det:\operatorname{GL}_n(-)\Rightarrow(-)^{\times} from commutative rings to groups

Example

For a fixed natural number n1n\ge1, entrywise application of ring homomorphisms makes invertible matrices and units group-valued functors, and determinant is natural between them.

Facts & Assumptions

Given: A natural number n1n\ge1 and unit-preserving homomorphisms of commutative rings.

Verification

technique · direct
1.1

For a commutative ring RR, put U(R)=R×U(R)=R^\times. For φ:RS\varphi:R\to S, restrict φ\varphi to units; it is a group homomorphism because ring homomorphisms preserve products, identities, and inverses. Identity and composition are inherited, so UU is a functor to Grp\mathbf{Grp}.

L1L2
1.2

Put Gn(R)=GLn(R)G_n(R)=\operatorname{GL}_n(R) and apply φ\varphi entrywise. From the product formula, φ((AB)ik)=jφ(aij)φ(bjk)\varphi((AB)_{ik})=\sum_j\varphi(a_{ij})\varphi(b_{jk}), so this assignment preserves matrix products and identities and carries an inverse matrix to an inverse matrix. Entrywise identity and composition make GnG_n a functor to Grp\mathbf{Grp}.

L1L2L3algebra
2.1

Multiplicativity and the unit result in [L4] make detR:Gn(R)U(R)\det_R:G_n(R)\to U(R) a group homomorphism.

step 1.1step 1.2L4
2.2

Applying φ\varphi to the finite Leibniz sum term by term gives detS(Gn(φ)(A))=φ(detR(A))=U(φ)(detR(A))\det_S(G_n(\varphi)(A))=\varphi(\det_R(A))=U(\varphi)(\det_R(A)).

step 1.1step 1.2L2L4
3.1

Step 2.2 is the naturality square for every commutative-ring homomorphism. Hence (detR)R(\det_R)_R defines a natural transformation GnUG_n\Rightarrow U.

step 2.1step 2.2L1

Depends on

Used by

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Dependency tree · next 3 levels

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Sources