@BalarkaSen Sorry in case I'm missing the point, but I think there's an easier way: Let $K/F$ be a finite extension and a splitting field for a polynomial over $F$. Let $f\in F[x]$ have a root $\alpha\in K$. Consider an irreducible factor $g$ of $f$ in $K[x]$. Consider the extension $K(\beta)/K$ by adjoining a root $\beta$ of $g$. $\alpha\mapsto\beta$ defines an $F$-embedding $F(\alpha)\rightarrow K(\beta)$.
This can be extended to an embedding $K\rightarrow L$, where $K(\beta)\subseteq L$. However, $K$ is is a splitting field, so the image of $K$ under this embedding is contained in $K$ again; in particular, $\beta\in K$, so $\deg g=[K(\beta)\colon K]=1$, hence $f$ splits completely.
Essentially, having a root gives room to embed, but all your embeddings are necessarily automorphisms, so you already have all the roots.
Interestingly (or not), I put this as a (not too hard) exam problem the last time I taught algebra: If $K/F$ is Galois and the irreducible polynomial $f(x)\in F[x]$ has a root in $K$, then it splits in $K$.
Oh, and I put it again on the final. Hmm ... I guess I like that result. :) Symmetry and all.
So, on a normed algebra $A$ of finite dimension, if $x_n \to x$ then $a x_n \to ax$ where $a \in A$, only because $x \mapsto ax$ is linear and hence continuous.. I'm asking for an example in infinite dimension where this breaks.
$x \mapsto ax$ is still continuous in infinite dimension because it is bounded (its operator norm is $\|a\|$)
@Thorgott "easier" is in the eye of the beholder
your proof is essentially the same, only that he is more explicit (he explained why "the image of $K$ under this embedding is contained in $K$ again" and he used an algebraic closure instead of some finite extension $L$
I think working with finite extensions is easier than working with the algebraic closure in this case, because you don't need choice, which makes it a little less technical. Of course, you are right that this is up to the beholder and that the general idea behind the proofs is similar.
Well, @Rithaniel, that's certainly false. If $n$ is even, negative the identity will work. But any upper (or lower) triangular matrix with $\pm 1$ on the diagonal will work.
Well, that's if we're looking at upper triangular or diagonal matrices. Do we know for sure that those are the only matrices in $SL_n(\mathbb{Z})$ that have only integer eigenvalues?
So, if $\frac{a}{b}$ is a root for $p(x)$ then $b$ divides the constant term and $a$ divides the leading term. Since we are assuming that $\frac{a}{b}\in\mathbb{Z}$, then we know that $b=\pm1$ and since the leading coefficient is $1$ we know that $a=\pm1$ and so $\frac{a}{b}=\pm1$ (This isn't what you asked me to prove, but it's where my mind went and seems relevant)
But couldn't $p(x)$ have no roots and still have a constant term? Such as $x^2+1$? Though, I suppose it has roots in the complex numbers and the product of those complex roots is 1.
@Tanuj My first inclination would be to add circuitry to handle initializing to 5 and then force a reset at 15. But that isn't exactly elegant. Are you looking for something cleaner? Or would this work? -Rithaniel's friend who is creeping over his shoulder.
@Tanuj Look at how you set each output bit (I'm assuming you're using JK flipflops). override the input to each JK when you reinitialize and force each bit to whatever you need it to be to get 5. -Not TedShifrin or Rithaniel
well it was riddled with errors and I posted here like three times before I posted a question about it and nobody responded which could of prevented the embarrassing public blunder gfc what ever happened to team adam
