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In lecture notes I am reading for Abstract Algebra, in the Separability section for Field Theory, it gives the introduction and definition that a separable polynomial is one that has no repeated roots in a splitting field. Then it gives an example of a separable polynomial by saying

Thus if $f(X) = (X-1)^2(X-3)$ over $\mathbb{Q}$, then $f$ is separable, because the irreducible factors $(X-1)$ and $(X-3)$ do not have repeated roots.

Is this correct? Isn't $(X-1)^2$ a repeated root at $X = 1$?

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  • $\begingroup$ the polynomial has a repeated root, but the irreducible factors do not $\endgroup$ – J. W. Tanner May 27 '20 at 17:41
  • $\begingroup$ @J.W.Tanner What would an irreducible factor with a repeated root look like $\endgroup$ – Ryan Shesler May 27 '20 at 17:43
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You are correct that $1$ is a repeated root.

Some authors use an older definition of separability which is that if $F$ is a field and $f(x)=kp_1(x)p_2(x)\dots p_n(x)$ is any polynomial $\in F[X]$ where the $p_i$s are all the monic irreducible factors of $f$, then $f$ is separable if all $p_i$s are separable, i.e. all $p_i$s have no repeated roots in their respective splitting fields/ in an algebraic closure of $F$ (bear in mind, the $p_i$s are taken to be irreducible over $F$ specifically).

The more modern definition is that a polynomial $f\in F[X]$ is separable if it has no repeated roots in its splitting field/ in an algebraic closure of $F$.

With more context (like the exact words of the original definition or further examples by the authors) you should be able to tell which is being used (although, based on the example given, I'd guess the older definition is being used by the authors here).

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    $\begingroup$ I get it. I suppose the point of confusion for me was that they chose an example where each irreducible factor split over the original field and did not need an extension. I found an example of an inseparable polynomial and your answer helped. Thank you. $\endgroup$ – Ryan Shesler May 27 '20 at 18:00

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