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"more than two elements" was irrelevant.
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Jeremy Rickard
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Even for a $2$-dimensional vector space over a field $\mathbb{F}$ with more than two elements (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.

Both $$\left\{\begin{pmatrix}x&0\\0&y\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ and $$\left\{\begin{pmatrix}x&y\\0&x\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ are maximal commutative subrings, but they're not isomorphic.

Even for a $2$-dimensional vector space over a field $\mathbb{F}$ with more than two elements (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.

Both $$\left\{\begin{pmatrix}x&0\\0&y\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ and $$\left\{\begin{pmatrix}x&y\\0&x\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ are maximal commutative subrings, but they're not isomorphic.

Even for a $2$-dimensional vector space (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.

Both $$\left\{\begin{pmatrix}x&0\\0&y\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ and $$\left\{\begin{pmatrix}x&y\\0&x\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ are maximal commutative subrings, but they're not isomorphic.

Source Link
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

Even for a $2$-dimensional vector space over a field $\mathbb{F}$ with more than two elements (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.

Both $$\left\{\begin{pmatrix}x&0\\0&y\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ and $$\left\{\begin{pmatrix}x&y\\0&x\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ are maximal commutative subrings, but they're not isomorphic.