Quotients of polynomial rings #
For a commutative ring $R$, evaluating a polynomial at an element $x \in R$ induces an isomorphism of $R$-algebras $R[X] / \langle X - x \rangle \cong R$.
Equations
- Polynomial.quotientSpanXSubCAlgEquiv x = (Ideal.quotientEquivAlgOfEq R ⋯).trans { toEquiv := (RingHom.quotientKerEquivOfRightInverse ⋯).toEquiv, map_mul' := ⋯, map_add' := ⋯, commutes' := ⋯ }
Instances For
For a commutative ring $R$, evaluating a polynomial at an element $y \in R$ induces an isomorphism of $R$-algebras $R[X] / \langle x, X - y \rangle \cong R / \langle x \rangle$.
Equations
- One or more equations did not get rendered due to their size.
Instances For
For a commutative ring $R$, evaluating a polynomial at elements $y(X) \in R[X]$ and $x \in R$ induces an isomorphism of $R$-algebras $R[X, Y] / \langle X - x, Y - y(X) \rangle \cong R$.
Equations
- One or more equations did not get rendered due to their size.
Instances For
If I is an ideal of R, then the ring polynomials over the quotient ring I.quotient is
isomorphic to the quotient of R[X] by the ideal map C I,
where map C I contains exactly the polynomials whose coefficients all lie in I.
Equations
- One or more equations did not get rendered due to their size.
Instances For
If P is a prime ideal of R, then R[x]/(P) is an integral domain.
Given any ring R and an ideal I of R[X], we get a map R → R[x] → R[x]/I.
If we let R be the image of R in R[x]/I then we also have a map R[x] → R'[x].
In particular we can map I across this map, to get I' and a new map R' → R'[x] → R'[x]/I.
This theorem shows I' will not contain any non-zero constant polynomials.
Given a domain R, if R[X] is a principal ideal ring, then R is a field.
Split off from quotientEquivQuotientMvPolynomial for speed.
Split off from quotientEquivQuotientMvPolynomial for speed.
If I is an ideal of R, then the ring MvPolynomial σ I.quotient is isomorphic as an
R-algebra to the quotient of MvPolynomial σ R by the ideal generated by I.
Equations
- One or more equations did not get rendered due to their size.