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Can any one help me what it means a polynomial is irreducible over function field over finite field with one example. Thank you very much.

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    $\begingroup$ $f \in K[t]$ means $f(t) = \sum_{n=0}^d a_n t^n, a_n \in K$. It is irreducible if it doesn't factor as $f = gh$ with $g,h \in K[t],0 < \deg(g) < \deg(f)$. Here you'll have $K = \mathbf{F}_p(x)$. $\endgroup$ – reuns Dec 9 '17 at 17:20
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Let $\mathbb F_5$ be the field of order $5$ and let $K=\mathbb F_5(X)$ be the function field over $\mathbb F_5$. Consider the polynomial $$P(T)=T^2 - \dfrac{1}{X}\in K[T]$$ with coefficients in $K$.

$P$ is irreducible over $K$ if and only if $P$ has no roots in $K$ (since $P$ is degree $2$).

Let us see by contradiction that $P$ is irreducible in $K$. Assume that $P$ has a root $\alpha$ in $K$. Then there are polynomials $Q(X), R(X)$ with coefficients in $\mathbb F_5$ so that $\alpha = \dfrac{Q(X)}{R(X)}$ and $\alpha^2 = \dfrac{1}{X}$. Equivalently $$ X\cdot Q(X) ^2 = R(X)^2 $$ as an equality in $\mathbb F_5[X]$. Left hand side has odd degree and right hand side has even degree.

Edit: Maybe it's important to remark that this example makes no use of properties of $\mathbb F_5$ other than being a field. So formally speaking this doesn't show any of the cool things of finites fields or positive characteristic fields.

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  • $\begingroup$ Thank you very much. One more doubt, what are the coefficients in K? Is it $F_5$ along with $X$ or only $F_5$ elements $\endgroup$ – thanks Dec 11 '17 at 4:43
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    $\begingroup$ K is the quotient field of $\mathbb F_5[X]$. That is elements of K are of the form P(X)/Q(X) with P, Q polynomials with coefficients in $\mathbb F_5$. $\endgroup$ – eduard Dec 11 '17 at 12:03
  • $\begingroup$ Might be this is silly, in the above example, how will I check is $P $is irreducible over $K[T]$ $\endgroup$ – thanks Mar 22 '18 at 12:22

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