In spite of the fact that the matrix ring $\mathbb{C}^{n \times n}$ is not a field, is it still possible to talk about it being 'algebraically closed' in the sense that $\forall f \in \mathbb{C}^{n \times n}[x]$ does $\exists A \in \mathbb{C}^{n \times n}$ such that $f(A) = 0$? If so, then is it 'algebraically closed'?

Are there any other non-field sets that this idea can be extended to?

  • 6
    $\begingroup$ The real question is here what $\mathbb{C}^{n\times n}\left[x\right]$ means. With $\mathbb{C}^{n\times n}$ being non-commutative, it is not clear what kind of polynomials you wish to allow. For instance, the Amitsur-Levitzki theorem ( gilkalai.wordpress.com/2009/05/12/… ) gives a nontrivial polynomial relation between any $2n$ matrices in $\mathbb{C}^{n\times n}$; if you replace the right hand side by $1$ rather than $0$, you will get a polynomial that never attains zero (though it is hardly the zero polynomial.) $\endgroup$ Feb 1, 2010 at 10:29
  • $\begingroup$ On the other hand, Ore polynomials (coefficients on the left, powers of $x$ on the right) may have a chance of making the assertion correct. $\endgroup$ Feb 1, 2010 at 10:30
  • $\begingroup$ Ok, suppose we define $f \in \mathbb{C}^{n \times n}[x]$ by $f(x) = \sum_{i=0}^n A_i x^i$ where $A_i \in \mathbb{C}^{n \times n}$, so we do in fact have coefficients on the left, powers of x on the right (this way infact what I was thinking of). Does this help or is there a better definition to use? I would imagine the simplest way to find a counter example would be finding a matrix with no 'square root', i.e. an $A \in \mathbb{C}^{n \times n}$ such that $\forall B \in \mathbb{C}^{n \times n}$, $B^2 \neq A$. $\endgroup$
    – Mark Bell
    Feb 1, 2010 at 10:57
  • 5
    $\begingroup$ Yuck. People actually call the ring of matrices $\mathbb C^{n\times n}$? When I see that, I assume you mean something like $\prod^{n^2} \mathbb C$, the commutative product of rings. Much better you be something like $\text{End}(\mathbb C^n)$. $\endgroup$ Feb 1, 2010 at 16:22
  • 5
    $\begingroup$ There's a model-theoretic generalization of "algebraically closed", namely "existentially closed": see Wikipedia. It basically says that if some system of equations has a solution in a larger algebra, then it has a solution within the algebra. Douglas' comment is that some nilpotent matrix in $M_2(C)$ has no square root. Embedding $M_2(C)$ into $M_4(C)$ diagonally, it has a square root in the larger algebra. So $M_2(C)$ is not existentially closed. $\endgroup$
    – YCor
    Jan 30, 2020 at 12:33

2 Answers 2


The matrix $\left( \begin{array}{cc} 0 & 1 \\\\ 0 & 0 \end{array} \right)$ has no square root.

Polynomials make sense for continuous complex functions on a space. If that space is $\mathbb R$, then polynomial equations with complex coefficients are solvable. If that space is $\mathbb C$ or $S^1$ then $g^2 = f$ may not be solvable.

  • 2
    $\begingroup$ And so $f(x) = \left(\begin{array}{cc}1 & 0 \\ 0 & 1 \\ \end{array}\right) x^2 - \left(\begin{array}{cc}0 & 1 \\ 0 & 0 \\ \end{array}\right) = 0$ has no root over $\mathbb{C}^{n \times n}$. Hence $\mathbb{C}^{n \times n}[x]$ is not 'algebraically closed'. $\endgroup$
    – Mark Bell
    Feb 1, 2010 at 11:52
  • 1
    $\begingroup$ What if you consider H = n x n Hermitian matrices? Does the spectral theorem provide the necessary push in this context? $\endgroup$ Feb 1, 2010 at 16:26
  • $\begingroup$ @Tom: do you mean allowing the polynomial to have coefficients in H? $\endgroup$
    – Yemon Choi
    Feb 1, 2010 at 19:48

I'd like to add that a nice theory of roots of polynomials over noncommutative rings was developed by I. Gelfand, V. Retakh, and R. Wilson, see the paper arXiv:math/0208146 and references therein (in particular, the earlier paper by Gelfand and Retakh on the noncommutatoive Vieta theorem).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .