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Update lecture18.tex #9

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2 changes: 1 addition & 1 deletion Lectures/lecture18.tex
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Expand Up @@ -333,7 +333,7 @@ \subsection{Идеальный спектр}
Мы знаем, что $0 = f_\text{min}(\varphi) = p(\varphi) h(\varphi)$.
Предположим, что $p(\varphi)$ обратим.
Тогда в равенстве $p(\varphi)h(\varphi) = 0$ можно сократить на $p(\varphi)$.
Значит $h(\varphi) = 0$, что противоречит минимальности $\varphi$.
Значит $h(\varphi) = 0$, что противоречит минимальности степени $f_\text{min}$.
\end{proof}


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