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Let $F$ be a field and $f(x)$ be an irreducible polynomial in $F[x]$. Show that the set of polynomials modulo $f(x)$ forms a field.

the set of polynomials modulo $ f(x)$: $r(x)=p(x)g(x)+f(x)$

Fields: with the two binary operations $+/\times$ and the distribution law: $(a+b)c=ac+bc$

Jyrki Lahtonen
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stedmoaoa
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  • Do you understand what "the set of polynomials modulo $f(x)$" means, and what are the operations there? Which part of the definition of a field are you having trouble with? – Robert Israel May 23 '17 at 17:39
  • When f(x) in the field, then how to should the set of $r(x)$ satisfies the +/x operation? – stedmoaoa May 23 '17 at 17:47
  • $F[x]/(f(x))$ is by definition a commutative ring. That the multiplication is inversible follows from the Euclid algorithm in $F[x]$, long division. – reuns May 23 '17 at 18:00
  • To be a field a structure must not only have addition and multiplication but also additive and multiplicative inverses. But contrast, a ring need not have multiplicative inverses, but still has multiplication and addition and additive inverses and the distributive law. – Michael Hardy May 23 '17 at 18:02
  • How to show the set of polynomials has addition and multiplication? – stedmoaoa May 23 '17 at 18:03
  • @MichaelHardy If $R$ is a commutative ring and $I$ an ideal of $R$ then $R/I$ is the commutative ring whose elements are subsets of $R$ of the form $a+I$ – reuns May 23 '17 at 18:11
  • @stedmoaoa You should start by showing that $\mathbb{C}$ is "really the same" as $\mathbb{R}[x]/(x^2+1)$. This generalizes without problem to $F[x]/(f(x))$ for any polynomial $f \in F[x]$ and field $F$. For showing it is a field whenever $f$ is irreducible, you'll need to look at the Bezout identity, Euclid algorithm, long division carefully. – reuns May 23 '17 at 18:14

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Most of the answer is showing that in the quotient ring of the ring of ring of polynomials modulo the ideal generated by $f(x),$ there are multiplicative inverses. One can show that be seeing how to actually find the multiplicative inverse of any non-zero element (and "non-zero" means in effect corresponding to any polynomial that is not a multiple of $f(x),$ i.e. is not $f(x)$ multiplied by some other polynomial.

The division algorithm applied to polynomials is this: Divide a polynomial $g(x)$ by a polynomial $h(x)$ and get a quotient and a remainder, and the degree of the remainder is less than the degree of the $h(x).$

The division algorithm is used in Euclid's algorithm. The way of finding multiplicative inverses by using Euclid's algorithm makes use of the quotients found when applying Euclid's algorithm. In this answer I explained how to do that with integers as opposed to polynomials. It's done the same way with polynomials. In order for this to work in every case, it it necessary that the polynomial $f(x)$ be irreducible.

Michael Hardy
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