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Let $R$ be a ring with identity (not necessarily commutative) and $R[x]$ be a ring of polynomials over $R$. We say that a ring $S$ is an extension of $R$ if there is a subring $\tilde{R}$ in $S$ isomorphic to $R$. Let $S$ be an extension of $R$, and $$\phi: R\to \tilde{R}\subset S$$ be a ring isomorphism. We say that a polynomial $f(x) = \sum\limits_{j\geq 0}f_jx^j\in R[x]$ has a root $\alpha\in S$ if $$ \sum\limits_{j\geq 0}\phi(f_j)\alpha^j = 0. $$

In the case, where $R$ is a commutative ring every monic polynomaial $f(x)\in R[x]$ has a root $[x]_f$ in the extension $S = R[x]/R[x]f(x)$ of $R$. In the case, where $R$ is not commutative the set $R[x]/R[x]f(x)$ is a left $R[x]$-module but not a ring, because an ideal $R[x]f(x)$ is not two-sided ideal, but only one-sided. Also in non-commutative case there are examples such that two-sided ideal, containing $f(x)$ that is an ideal $R[x]f(x)R[x]$ is equal to $R[x]$ and in this case $R[x]/R[x]f(x)R[x]$ isomorphic to zero ring.

I want to prove that for every ring with identity $R$ and every monic polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

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  • $\begingroup$ How could $R[x]f(x)R[x]$ be equal to $R[x]$ when the degree of $f$ is not 0? $\endgroup$ – Uri Bader Apr 30 '16 at 15:56
  • $\begingroup$ @user89334 The example in my answer below does just that. The ideal generated by $1+ax$ contains $1=(1-ax)(1+ax)$. $\endgroup$ – Pace Nielsen Apr 30 '16 at 16:11
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    $\begingroup$ In general, even in the commutative case, polynomials don't have roots in an extension. Example: $f=1 + 2x\in (\mathbb{Z}/4)[x]$. The problem is that $R[x]/(f)$ is in general no extension of $R$. $\endgroup$ – Todd Leason Apr 30 '16 at 16:12
  • $\begingroup$ My comment was very stupid. I came back to erase it, but figure its too late... $\endgroup$ – Uri Bader Apr 30 '16 at 16:22
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    $\begingroup$ Note: "unitary polynomial" means "monic polynomial". $\endgroup$ – YCor Apr 30 '16 at 16:29
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In the noncommutative case, your condition for a "root" is called a "right root". I remember that T.Y. Lam worked with this condition a bit (you might search through his papers, or look in his "First Course in Noncommutative Rings" book).

It is easy to get commutative rings where your condition fails (because even though $R[x]f(x)$ is an ideal in $R[x]$, it still intersects $R$)! For instance, let $F$ be a non-zero ring with identity and consider the ring

$$R=F[a\ :\ a^2=0].$$

The polynomial $f(x)=1+ax\in R[x]$ cannot have any root in any extension ring of $R$, because this would force $a$ to be a unit in the extension ring, but also nilpotent.


Edited to add: I've been thinking about the new question and monic polynomials. Without loss of generality think of $R$ as contained in $S$. If $R$ is allowed to have a different unit than $S$, I think that the answer is positive. When $R$ is forced to be a unital subring of $S$ (i.e. containing the same unit) the answer gets a bit harder, as I'll describe below.

In the latter case, take $S:=R\coprod_{\mathbb{Z}}\mathbb{Z}[t]/(f(t))$. Our goal is to show that $R$ is a unital subring of $S$, and we are done. We can do that by writing elements of $S$ in reduced form.

To explain the motivation, take $f$ to be a quadratic polynomial for a moment. Say $f(x)=x^2+bx+c$ with $b,c\in R$. Ignoring the "bar" notation (since $S$ is a factor ring) for convenience, we have the relation

$$t^2\mapsto -bt-c.$$

We can reduce $t^3$ in two ways, and that gives us a new relation

$$tbt\mapsto -bt-tc+ct-bc.$$

We can now reduce $t^2bt$, $tbt^2$, and $tbtbt$ in two ways (each), and we get another relation beginning $tb^2t\mapsto\cdots$. As long as $b^n\notin \mathbb{Z}$ for each $n\geq 1$, I believe that this demonstrates that your question is positively answered (for quadratics, and this can be extended).

When $b^n\in \mathbb{Z}$ things get more complicated, but the problem may still be tractable.

When $R$ is allowed to have a different identity from $S$, there is an even easier construction.

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    $\begingroup$ I found the simpler example in the commutative case and beat you for 2 minutes. -:) $\endgroup$ – Todd Leason Apr 30 '16 at 16:15
  • $\begingroup$ @Pace Nielsen, Thank you very much, but I'm interesting only in the unitary polynomials. I will clarify my question. $\endgroup$ – Mikhail Goltvanitsa Apr 30 '16 at 16:24
  • $\begingroup$ @Pace Nielsen, Thank you. Can you explain more circumstantially what is the construction of $R\coprod_{\mathbb{Z}}\mathbb{Z}[t]/(f(t))$. Also I don't understand, why $t$ don't commute with elements of $R$. $\endgroup$ – Mikhail Goltvanitsa May 4 '16 at 14:12
  • $\begingroup$ @MikhailGoltvanitsa The elements of $R$ don't commute with $t$, because this is the coproduct of rings. For instance the ring $\mathbb{Z}[s]\coprod_{\mathbb{Z}}\mathbb{Z}[t] = \mathbb{Z}\langle s,t\rangle$ has $s,t$ non-commuting. I can't explain the intricacies of the coproduct of rings in this comment, so I recommend finding a good book on the subject. $\endgroup$ – Pace Nielsen May 4 '16 at 14:35
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I have found a very simple and demonstrative construction of the ring extension, which came from non-commutative generalization of Hamilton-Caley's Theorem.

Let $R$ be a ring and $f(x) = x^m-\sum\limits_{j=0}^{m-1}f_jx^j\in R[x]$ be a monic polynomial. We identify ring $R$ with a subring $\tilde{R} = \{\mathrm{diag}(r,r,\ldots,r): r\in R\}\subset M_m(R)$.

Then $f(x)$ has a root of the form $$ \alpha=\left(\begin{array}{llllll} 0& e& 0&\ldots& 0& 0 \\ 0& 0& e&\ldots& 0& 0 \\ . & . & . & . & .& . \\ 0& 0&0&\ldots& 0& e \\ f_0& f_1& f_2&\ldots& f_{m-2}& f_{m-1} \\ \end{array} \right) $$ That is $\alpha^m-\sum\limits_{j=0}^{m-1}f_j\alpha^j = 0\in M_m(R)$. But we note that in general: $f(\alpha^T)\neq 0$.

See article for proof of non-commutative generalization of Hamilton-Caley's Theorem.

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    $\begingroup$ Nice! Note that your properties $(*)$ and $(**)$ are what Caracciolo, Sportiello and Sokal (in arXiv:0809.3516v2) call "column-commutativity" and "row-commutativity". $\endgroup$ – darij grinberg Jul 29 '17 at 20:01
  • $\begingroup$ Mikhail, looking at the reference you give, the proof of the non-commutative generalization of the Caley-Hamilton theorem is done only for Galois rings (which are finite commutative rings). Finding roots over arbitrary commutative rings is trivial, and doesn't need to pass through matrix rings. $\endgroup$ – Pace Nielsen Aug 1 '17 at 18:52
  • $\begingroup$ I tried looking for a good source for the fact that a companion matrix of a monic polynomial satisfies that polynomial (even over a noncommutative ring). Apparently it is just a "classical fact" that nobody has bothered to state. The only direct reference I could find was in a paper of Andre Leroy. $\endgroup$ – Pace Nielsen Aug 2 '17 at 17:00
  • $\begingroup$ @PaceNielsen, in the given refference the proof of non-commutative Hamilton-Caley Theorem is given for arbitrary non-commutative ring with identity. $\endgroup$ – Mikhail Goltvanitsa Aug 20 '17 at 10:38

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