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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 $S$.

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 polynomial $f(x)\in R[x]$ 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 the 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, the ideal $R[x]f(x)R[x]$, is equal to $R[x]$, and in this case $R[x]/R[x]f(x)R[x]$ is isomorphic to zero ring.

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

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 unitary 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 unitary polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

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 $S$.

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 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 polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

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 unitary 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 unitary polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.