Proof that complex roots of a polynomial occur in pairs

[sto]

Let R be the field of real numbers and C be the field of complex numbers.
Let p:R-->R be a polynomial (with real coefficients).

I want to prove that if a complex number c is a root of p, then so is
its complex conjugate c* (i.e. the complex roots of a polynomial occur
in pairs).

Consider C as an R-algebra. If a,b in C, then because (a+b)* = a*+b*
and (ab)* = a*b*, the map c-->c* is an isomorphism of algebras. It
follows that p(c)-->p(c*). Since by assumption p(c)=0, and because any
isomorphism will map the identity element (zero) in one space to the
identity element in the other space, it follows that p(c*)=0.

Is this proof rigorous and correct?

Thanks,
-sto

[Chip Eastham]

Post by sto
Let R be the field of real numbers and C be the field of complex numbers.
Let p:R-->R be a polynomial (with real coefficients).
I want to prove that if a complex number c is a root of p, then so is
its complex conjugate c* (i.e. the complex roots of a polynomial occur
in pairs).
Consider C as an R-algebra.  If a,b in C, then because (a+b)* = a*+b*
and (ab)* = a*b*, the map c-->c* is an isomorphism of algebras. It
follows that p(c)-->p(c*).  Since by assumption p(c)=0, and because any
isomorphism will map the identity element (zero) in one space to the
identity element in the other space, it follows that p(c*)=0.
Is this proof rigorous and correct?
Thanks,
-sto

Yes, though I'd stress that complex conjugation
fixes all real numbers, not just zero.

--c

[Axel Vogt]

Post by sto
Let R be the field of real numbers and C be the field of complex numbers.
Let p:R-->R be a polynomial (with real coefficients).
I want to prove that if a complex number c is a root of p, then so is
its complex conjugate c* (i.e. the complex roots of a polynomial occur
in pairs).
Consider C as an R-algebra. If a,b in C, then because (a+b)* = a*+b*
and (ab)* = a*b*, the map c-->c* is an isomorphism of algebras. It
follows that p(c)-->p(c*). Since by assumption p(c)=0, and because any
isomorphism will map the identity element (zero) in one space to the
identity element in the other space, it follows that p(c*)=0.
Is this proof rigorous and correct?
Thanks,
-sto

I would prefer to extend the iso to the polynomial algebras and
see it then to commute with evaluation (= polynomials viewed as
functions).

[William Elliot]

Post by sto
Let R be the field of real numbers and C be the field of complex numbers.
Let p:R-->R be a polynomial (with real coefficients).
I want to prove that if a complex number c is a root of p, then so is its
complex conjugate c* (i.e. the complex roots of a polynomial occur in pairs).
Consider C as an R-algebra. If a,b in C, then because (a+b)* = a*+b* and
(ab)* = a*b*, the map c-->c* is an isomorphism of algebras. It follows that
p(c)-->p(c*). Since by assumption p(c)=0, and because any isomorphism will
map the identity element (zero) in one space to the identity element in the
other space, it follows that p(c*)=0.

Too complicated. Assume p in R[x], p(a) = 0.
Show p(a*) = (p(a))^*. Thus p(a*) = 0* = 0.

Post by sto
Is this proof rigorous and correct?
Thanks,
-sto