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If $\alpha$ is an algebraic element of $\mathbb{C}$, then there is a unique non-zero polynomial $f \in \mathbb{Q}[x]$ with leading coefficient $1$ such that $f(\alpha) = 0$, and $f$ is irreducible.

The first part of this proof would be proving that $f$ is not a unit, but what does the concept of a unit mean in the set of polynomials? I can't see how a polynomial would have a $2$ sided inverse under multiplication?

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Constants are also polynomials; they are polynomials of degree 0. The polynomial 1 is a unit because it has an inverse. –  MJD May 20 '12 at 13:09
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up vote 2 down vote accepted

A unit in the ring of polynomials over some structure (say, a field) is the same as in any other ring: an element that has a multiplicative inverse within that ring.

If you want to focus on polynomials over fields then the units in that ring are precisely the constant polynomials $\,f(x)=k\,,\,0\neq k\,$ a constant.

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The second paragraph is true iff the coefficient ring is reduced (i.e. only $0$ is nilpotent), see my answer. –  Bill Dubuque May 20 '12 at 14:23
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It would, if it were of degree zero.

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So all polynomials of degree greater than 0 are non-units? –  user26069 May 20 '12 at 13:13
    
In the ring you are talking about, yes. –  Gerry Myerson May 20 '12 at 13:23
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Hint $\rm\:\ f\:\!$ unit (i.e. invertible) $\rm\: \Rightarrow\: 1 = f(x)g(x)\:\Rightarrow\: 1 = f(\alpha)g(\alpha) = 0\cdot g(\alpha) = 0$

The special case $\rm\:f(x) = x\:$ is one of my most popular posts (due to its universal view?)

Generally $\rm\:\sum a_i\:\!x^i\:$ is a unit in $\rm R[x] \iff a_0\:$ is a unit$\rm\:R\:$ and $\rm\:a_1,...,a_n$ are nilpotent in $\rm R.$

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