ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Multiplication by $2$ on $Z$ is injective but not surjective: its determinant is the non-unit $2$ , its adjugate is integral, and its inverse exists after extending scalars to $Q$

Example

The coordinate endomorphism $T_{[2]} : Z \to Z$ is injective but not surjective. Its determinant is the non-unit $2$ , its adjugate is $[1]$ , and after extending scalars to $Q$ its inverse is $[1 / 2]$ .

Facts & Assumptions

Given: The $1 \times 1$ matrix $A = [2]$ .

[F1]

$Z$ is a commutative ring (The integers form a commutative ring) and its only units are $1$ and $- 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L1]

Integer multiplication has cancellation (The integers have no zero divisors; multiplicative cancellation).

[F2]

$T_{A} (x) = A x$ , $\det (T_{A}) = \det (A)$ , and $adj (T_{A}) = T_{adj (A)}$ (For $A \in M_{n} (R)$ , the coordinate endomorphism $T_{A} : R^{n} \to R^{n}$ , with $\det (T_{A}) : = \det (A)$ and $adj (T_{A}) : = T_{adj (A)}$ ).

[L2]

A positive-sized square matrix over a commutative ring is invertible exactly when its determinant is a unit (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

[L3]

If the determinant is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[F3]

$Q$ is a field (The rationals form a field).

Verification

technique · direct

1.1

From the $1 \times 1$ definitions, $T_{A} (n) = 2 n$ , $\det (A) = 2$ , and $adj (A) = [1]$ , because the unique empty minor has determinant $1$ .

F1F2algebra

1.2

If $2 m = 2 n$ , integer cancellation gives $m = n$ , so $T_{A}$ is injective. It is not surjective because $2 n = 1$ has no integer solution.

L1algebra

2.1

The element $2$ is not a unit of $Z$ by [F1], so [L2] agrees that $A$ is not invertible over $Z$ .

step 1.1F1L2

2.2

Over the field $Q$ , $2$ is a unit. Formula [L3] and step 1.1 give $A^{- 1} = 2^{- 1} [1] = [1 / 2]$ .

F3L3step 1.1

3.1

Steps 1.1 through 2.2 establish every claim.

step 1.1step 1.2step 2.1step 2.2∎

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 82 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

András Pál, Introduction to Commutative Algebra (standard reference, not scraped)

Multiplication by 22 on ZZ is injective but not surjective: its determinant is the non-unit 22, its adjugate is integral, and its inverse exists after extending scalars to QQ