Exercise · ZW-FIRST-ISOMORPHISM-THEOREM-FOR-RINGS-007 · ★★☆ Intermediate

Let \(\varphi\colon \mathbb{Z}_{6}\to \mathbb{Z}_{2}\times \mathbb{Z}_{3}\) be given by \(\varphi(x)=(x\bmod 2,\ x\bmod 3)\). What is its kernel?

Choose one option.

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  • A. \(\{0\}\)

Explanation

If \(x\) is even and a multiple of 3 in \(\mathbb{Z}_{6}\), then \(x=0\). So \(\varphi\) is injective; since both rings have 6 elements, it is an isomorphism.

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Review the concept: First isomorphism theorem for rings

If \(\varphi\colon R\to S\) is a ring homomorphism with kernel \(\ker\varphi\), the quotient ring of \(R\) by \(\ker\varphi\) is isomorphic to the image of \(\varphi\).

\[R/\ker\varphi\;\cong\;\varphi(R)\]

More practice problems: first isomorphism theorem for rings

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