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Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Suppose I have an expression $y + f(x)$, and I know $y in mathbb R, x in X$ and $f$ is an arbitrary function from $X$ to the reals. If I know $y$ and $x$ but don't know $f$, I can't evaluate the expression $y + f(x)$. At least, not completely. But once I know $f$, (say $f(z) = z^2$), then I can evaluate the expression $y + f(x)$. Is there a name for the difference between these two circumstances? computability symbolic-computation share | cite | improve this question asked Aug 6 at 23:26 alexpghayes 101 3 In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters? – Suzet Aug 6 at 23:53 @Suzet That is not true. You can easily define a relation, for wh

Clash Royale CLAN TAG #URR8PPP up vote 4 down vote favorite 1 Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$. Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$. Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$. Attempt $[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$. $[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible. $mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$. Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$. If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$. How can I claim it is not possible? Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$ galois-theory splitting-field share | cite | improve this question asked Aug 6 at 23:40 Jo' 14