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Let $F$ be a field and let $f (x)$ be a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has a zero.

Since $F[x]$ is a unique factorization domain, $f (x)$ has an irreducible factor, say, $p(x)$. Clearly, it suffices to construct an extension field $E$ of $F$ in which $p(x)$ has a zero. Our candidate for $E$ is $F[x]/p(x)$. Due to a previous theorem, this is a field. Also, since the mapping of $\phi: F \rightarrow E$ given by $\phi(a) = a + \langle p(x) \rangle$ is one-to-one and preserves both operations, $E$ has a subfield isomorphic to $F$. We may think of $E$ as containing $F$ if we simply identify the coset $a + \langle p(x) \rangle$ with its unique coset representative a that belongs to $F$.

Then to show $p(x)$ has a zero in $E$, we substitute in $x + \langle p(x) \rangle$ into $p(x)$ and get $p(x) + \langle p(x) \rangle$ = $0 + \langle p(x) \rangle$.

My question is:

1. Why is $0 + \langle p(x) \rangle$ considered the $0$ of E and not the unity element of E?

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    $\begingroup$ Let $I=\langle p(x)\rangle$. Then $(0+I)(a+I)=0\cdot a+I=0+I$ for every $a\in E$. The unity is $1+I$ since $(1+I)(a+I)=1\cdot a+I=a+I$. $\endgroup$
    – Angelo Rendina
    Sep 18, 2016 at 14:37
  • $\begingroup$ Because $(0+I)(a+I)=0+I$, so cannot be a unit. The question is in fact not really on the topic of the title. $\endgroup$
    – Dietrich Burde
    Sep 18, 2016 at 14:38

1 Answer 1

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Becasue it is always the case that $a+I=0+I$ if $a\in I$ (and $I$ is an ideal of some ring)

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