Quotients of polynomial rings #
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- Polynomial.quotientSpanXSubCAlgEquivAux1 x = Ideal.quotientEquivAlgOfEq R (_ : Ideal.span {Polynomial.X - ↑Polynomial.C x} = RingHom.ker (Polynomial.evalRingHom x))
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For a commutative ring $R$, evaluating a polynomial at an element $x \in R$ induces an isomorphism of $R$-algebras $R[X] / \langle X - x \rangle \cong R$.
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For a commutative ring $R$, evaluating a polynomial at an element $y \in R$ induces an isomorphism of $R$-algebras $R[X] / \langle x, X - y \rangle \cong R / \langle x \rangle$.
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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 $R[X, Y] / \langle X - x, Y - y(X) \rangle \cong R$.
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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
.
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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.
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
.
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